Permutations and combinations — Paper 6 · A Level Mathematics 9709

3 Issam has 11 different CDs, of which 6 are pop music, 3 are jazz and 2 are classical. (i) How many different arrangements of all 11 CDs on a shelf are there if the jazz CDs are all next to each other? [3] (ii) Issam makes a selection of 2 pop music CDs, 2 jazz CDs and 1 classical CD. How many different possible selections can be made? [3]

6 marks

Mark scheme: 3 (i) 3! ×8!×9 M1 For k3! seen, k a +ve integer, accept 3P3 M1 For using m8! or n9! Seen, m and n +ve integers, accept m 8P8 etc = 2,177,280 or 2,180,000 A1 3 Correct final answer (ii) 6C2× 3C2× 2C1 M1 Multiplying 3 combinations or 3 numbers or 3 permutations together only B1 All of 6C2 and 3C2 and 2C1 seen (15, 3, 2) = 90 A1 3 Correct answer

This question in 9709/61 May/June 2008

4 A builder is planning to build 12 houses along one side of a road. He will build 2 houses in style A, 2 houses in style B, 3 houses in style C, 4 houses in style D and 1 house in style E. (i) Find the number of possible arrangements of these 12 houses. [2] (ii) Road First group Second group The 12 houses will be in two groups of 6 (see diagram). Find the number of possible arrangements if all the houses in styles A and D are in the first group and all the houses in styles B, C and E are in the second group. [3] (iii) Four of the 12 houses will be selected for a survey. Exactly one house must be in style B and exactly one house in style C. Find the number of ways in which these four houses can be selected. [2]

7 marks

Mark scheme: 12! 4 (i) = 831600 M1 Dividing by 3! 4! and 2! once or twice o.e !4!3!2!2 A1 [2] Correct final answer !6 !6 !6 !6 (ii) × B1 and seen o.e !2!4 !3!2 !2!4 !3!2 M1 multiplying their numbers for group 1 with their numbers for group 2 = 900 A1 [3] correct final answer (iii) 2 ×3 × 7C2 or 2 × 3 × 21 M1 7C2 seen multiplied or 5 options added = 126 A1 [2] correct final answer GCE A/AS LEVEL – October/November 2008 9709 06

This question in 9709/61 Oct/Nov 2008

4 A choir consists of 13 sopranos, 12 altos, 6 tenors and 7 basses. A group consisting of 10 sopranos, 9 altos, 4 tenors and 4 basses is to be chosen from the choir. (i) In how many different ways can the group be chosen? [2] (ii) In how many ways can the 10 chosen sopranos be arranged in a line if the 6 tallest stand next to each other? [3] (iii) The 4 tenors and 4 basses in the group stand in a single line with all the tenors next to each other and all the basses next to each other. How many possible arrangements are there if three of the tenors refuse to stand next to any of the basses? [3]

8 marks

Mark scheme: 4 (i) 13C10 × 12C9 × 6C4 × 7C4 M1 Expression involving the product of 4 combinations = 33033000 (33000000) A1 [2] Correct final answer allow 33×106 or 3.3×107 (ii) 5! × 6! B1 6! or 5! or 4! oe seen no denom = 86400 M1 a single product involving 6! and either 4! or 5! no denom A1 [3] Correct final answer (iii) 4! × 3! × 2 B1 4! or 3! or 4!/4 seen M1 a single product involving 3! (or 4!/4) and 4! = 288 A1 [3] Correct final answer

This question in 9709/61 May/June 2009

5 (a) Find how many numbers between 5000 and 6000 can be formed from the digits 1, 2, 3, 4, 5 and 6 (i) if no digits are repeated, [2] (ii) if repeated digits are allowed. [2] (b) Find the number of ways of choosing a school team of 5 pupils from 6 boys and 8 girls (i) if there are more girls than boys in the team, [4] (ii) if three of the boys are cousins and are either all in the team or all not in the team. [3]

11 marks

Mark scheme: 5 (a) (i) 1 × 5 × 4 × 3 or 5C3 × 3! or 5P3 M1 One of these oe = 60 A1 [2] Correct final answer (ii) 1 × 63 = 216 M1 Seeing 63 A1 [2] Correct answer (b) (i) 5G 0B = 8C5 = 56 (× 6C0) M1 Σ 2 or three 2-factor products, C or P 4G 1B = 8C4 × 6C1 = 420 B1 Any correct option unsimplified 3G 2B = 8C3 × 6C2 = 840 A1 A second correct option unsimplified total = 1316 A1 [4] Correct answer (ii) 11C2 + 11C5 M1 Adding two single perm or comb options 11Cx + 11Cy = 55 + 462 B1 One correct unsimplified option = 517 A1 Correct answer OR cousins in P(3B, 2G) + P(4B, 1G) M1 Σ 5 or more 2-factor perm or comb terms + P(5B, 0G) + cousins out P(3B, 2G) + P(2B, 3G) + P(1B, 4G) + P(0B, 5G) B1 3 or more correct unsimplified options = 28 + 24 + 3 + 28 + 168 + 210 + 56 = 517 A1 [3] Correct answer GCE A/AS LEVEL – October/November 2009 9709 61 4 C 2 × 7 C1 M1 U i 2 b lt f t d 1 b f

This question in 9709/61 Oct/Nov 2009

4 (a) (i) Find how many different four-digit numbers can be made using only the digits 1, 3, 5 and 6 with no digit being repeated. [1] (ii) Find how many different odd numbers greater than 500 can be made using some or all of the digits 1, 3, 5 and 6 with no digit being repeated. [4] (b) Six cards numbered 1, 2, 3, 4, 5, 6 are arranged randomly in a line. Find the probability that the cards numbered 4 and 5 are not next to each other. [3]

8 marks

Mark scheme: 4 (a) (i) 24 B1 [1] Correct final answer (ii) 3 digit odd 500+ = 4 ways M1 Attempt for 3 digit odd numbers 3 digit odd 600+ = 3 × 2 = 6 ways 4 digit odd 1000+ = 4 ways M1 Attempt for 4 digit odd numbers 4 digit odd 3000+ = 4 ways 4 digit odd 5000+ = 4 ways 4 digit odd 6000+ = 6 ways M1 For summing their number of ways with 3-digits OR 4 digit odd, last digit in 3 ways, and their number of ways with 4-digits 2nd to last in 3 ways, 2nd in 2 ways first in 1 way = 18 Total = 28 ways A1 [4] Correct total (b) no of ways 4 and 5 not next to each other = 6! − 5! × 2! = 720 − 240 M1 Finding ways digits not next to each other = 480 B1 240 or 480 seen Prob not next = 480/720 = 2/3 A1 [3] Correct answer

This question in 9709/62 Oct/Nov 2009

6 (i) Find the number of different ways that a set of 10 different mugs can be shared between Lucy and Monica if each receives an odd number of mugs. [3] (ii) Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a different design. Find in how many ways these 9 mugs can be arranged in a row if the china mugs are all separated from each other. [3] (iii) Another set consists of 3 identical red mugs, 4 identical blue mugs and 7 identical yellow mugs. These 14 mugs are placed in a row. Find how many different arrangements of the colours are possible if the red mugs are kept together. [3]

9 marks

Mark scheme: 6 (i) 10C1+ 10C3 + 10C5 + 10C7 + 10C9 M1 Summing some 10C combinations with odd numbers, all different A1 At least 3 correct unsimplified expressions = 512 A1 Correct answer [3] (ii) 6! × 7 × 6 × 5 B1 6! seen M1 multiplying by 7P3 o.e. = 151200 A1 correct answer [3] (iii) 12! / (4! × 7!) B1 12! Seen M1 dividing by 4!7! = 3960 A1 correct answer [3] GCE AS/A LEVEL – May/June 2010 9709 61 t

This question in 9709/61 May/June 2010

4 Three identical cans of cola, 2 identical cans of green tea and 2 identical cans of orange juice are arranged in a row. Calculate the number of arrangements if (i) the first and last cans in the row are the same type of drink, [3] (ii) the 3 cans of cola are all next to each other and the 2 cans of green tea are not next to each other. [5]

8 marks

Mark scheme: 4 (i) ends cola, 5!/2!2! = 30 M1 Considering all three options ends green tea, 5!/3!2! = 10 ends orange juice, 5!/3!2! = 10 A1 Any one option correct total = 50 ways A1 Correct answer 3 2 2 1 2 1 OR P(ends same) = × + × + × M1 OR Considering all three options 7 6 7 6 7 6 5 = A1 Correct fraction 21 5 !7 × = 50 ways A1 Correct answer 21 !2!2!3 [3] (ii) colas together, no restrictions, 5!/2!2! M1 Considering all colas together, or 5! seen = 30 ways A1 Correct answer colas together and green tea together, 4!/2! M1 Considering all colas tog and all green tea tog, or 4! seen = 12 ways A1 Correct answer 30 – 12 = 18 ways. A1 Correct final answer OR1 Attempt to list M1A1 OR1 10 or more, 12 or more correct M1A1 14 or more, 16 or more correct A1 18 correct 4 × 3 OR2 3 × = 18 M1 OR2 Considering all colas together, or 3! seen 2 A1 3 ways for colas and orange juice M1 Considering green teas not together A1 4 × 3 or (4 × 3)/2 A1 Correct final answer [5]

This question in 9709/63 May/June 2010

6 Pegs are to be placed in the four holes shown, one in each hole. The pegs come in different colours and pegs of the same colour are identical. Calculate how many different arrangements of coloured pegs in the four holes can be made using (i) 6 pegs, all of different colours, [1] (ii) 4 pegs consisting of 2 blue pegs, 1 orange peg and 1 yellow peg. [1] Beryl has 12 pegs consisting of 2 red, 2 blue, 2 green, 2 orange, 2 yellow and 2 black pegs. Calculate how many different arrangements of coloured pegs in the 4 holes Beryl can make using (iii) 4 different colours, [1] (iv) 3 different colours, [3] (v) any of her 12 pegs. [3]

9 marks

Mark scheme: 6 (i) 6P4 = 6!/2! = 360 B1 Correct answer [1] (ii) 4!/2! = 12 B1 Correct answer [1] (iii) 4! × 6C4 = 360 or 6P4 B1 Correct final answer [1] (iv) e.g. 2R 1B 1G, 1R 2B 1G, 1R 1B 2G M1 4!/2! seen !4 !4 !4 = + + = 36, mult by 6C3 M1 Mult by 6C3 !2 !2 !2 total = 720 A1 Correct answer [3] (v) 2R 2B = 4!/2!2! = 6 M1 Considering 2 colours e.g. RRBB or RBBR or... Mult by 6C2, total = 90 A1 mult by 6C2 Answer = 360 + 720 + 90 = 1170 A1ft Ft their (iii) + (iv) + (v) [3] GCE A LEVEL – October/November 2010 9709 61

This question in 9709/61 Oct/Nov 2010

6 Windows Back Front Aisle Windows A small aeroplane has 14 seats for passengers. The seats are arranged in 4 rows of 3 seats and a back row of 2 seats (see diagram). 12 passengers board the aeroplane. (i) How many possible seating arrangements are there for the 12 passengers? Give your answer correct to 3 significant figures. [2] These 12 passengers consist of 2 married couples (Mr and Mrs Lin and Mr and Mrs Brown), 5 students and 3 business people. (ii) The 3 business people sit in the front row. The 5 students each sit at a window seat. Mr and Mrs Lin sit in the same row on the same side of the aisle. Mr and Mrs Brown sit in another row on the same side of the aisle. How many possible seating arrangements are there? [4] (iii) If, instead, the 12 passengers are seated randomly, find the probability that Mrs Lin sits directly behind a student and Mrs Brown sits in the front row. [4]

10 marks

Mark scheme: 6 (i) 14P12 M1 14P12 seen oe = 4.36 × 1010 A1 Correct answer [2] (ii) business people 3! = 6 B1 3! oe seen, not in denominator students 5! = 120 B1 5! oe seen, not in denominator married couples 3P2 × 2 × 2 = 24 B1 24 oe seen, not in denominator total ways = 17280 B1 correct final answer [4] (iii) Mrs Brown 3 B1 any 2 of 3, 10, 5 oe seen, not in Mrs Lin 10 denominator Student 5 Prob = 3 × 10 × 5 × 11P9 / (i) B1 11P9 seen multiplied M1 dividing by their (i) = 0.0687 A1 correct answer [4] OR1 3/14 × 10/13 × 5/12 = 150/2184 (0.0687) B1 any 2 of numerators 3, 10, 5 oe seen B1 denominators 14, 13, 12 of 3 fractions M1 multiplying 3 separate fractions A1 correct answer OR2 1 − 3/14 = 11/14 B1 1 − 3/14 seen 1 − 11/14 × 5/13 = 127/182 B1 1 − 11/14 × 5/13 seen 8/14(4/13 × 12/12 + 9/13 × 7/12) + M1 attempt to find P(Mrs Lin not behind a 3/14(3/13 × 12/12 + 10/13 × 7/12) student and Mrs Brown not in front row), = 1206/2184 involving 8/14 × prob + 3/14 × prob 1 − (1524 + 1716 − 1206)/2184 = 150/2184 A1 correct answer GCE A LEVEL – October/November 2010 9709 63

This question in 9709/63 Oct/Nov 2010

2 Fahad has 4 different coloured pairs of shoes (white, red, blue and black), 3 different coloured pairs of jeans (blue, black and brown) and 7 different coloured tee shirts (red, orange, yellow, blue, green, white and purple). (i) Fahad chooses an outfit consisting of one pair of shoes, one pair of jeans and one tee shirt. How many different outfits can he choose? [1] (ii) How many different ways can Fahad arrange his 3 jeans and 7 tee shirts in a row if the two blue items are not next to each other? [2] Fahad also has 9 different books about sport. When he goes on holiday he chooses at least one of these books to take with him. (iii) How many different selections are there if he can take any number of books ranging from just one of them to all of them? [3]

6 marks

Mark scheme: Σx A 2 2 (ii) − 6.3 = 1.9252 M1 Attempt to find ΣxA using correct 9 variance formula ΣxA 2 = 150 A1 Correct ΣxA 2 1500. + 352 2 − .4017 = 4.780 M1 Using 352 + their 150 in correct variance 24 formula sd = 2.19 A1 [4] Correct answer 2 (i) 4 × 3 × 7 = 84 B1 [1] Correct answer (ii) 10! – 9! × 2 B1 10! − k × 9! seen oe = 2903040 (2900000) B1 [2] Correct answer OR 8! × 9 × 8 B1 8! × 9 × l seen oe = 2903040 (2900000) B1 Correct answer (iii) 9C1 + 9C2 + ... + 9C9 M1 Using combinations M1 Adding 9 combinations = 511 A1 [3] Correct answer OR 29 – 1 M1 29 seen M1 Subtracting 1 = 511 A1 Correct answer 3 (i) medianA < 35 or 20 ≤ medianA < 35 or B1 Correct numerical statement re medianA or medianA = 33.0/33.1/33.5/33.6 medianB or medianB ≥ 50 or 50 ≤ medianB < 70 or B1 [2] Correct numerical statement re other medianB = 51.7/51.9/52.2/52.4 median and a conclusion medianB > medianA OR A has 66 cand 50 < mark < 100, so medA < 50 B1 As before or B has 156 cand 50 < mark < 100, so medB > 50 medianB > medianA B1 As before (ii) 159 – 68 = 91 B1 [1] Correct final answer  5.4 × 25 + 14 5. × 43 + 27 × 91  (iii) mean=   / 300 M1 Using an attempt at mid-points, not end  + … + 84 5. × 40  points or class widths M1 Using an attempt at frequencies, not cum freqs M1 Sum of 6 prods, correct freqs, divided by 300 = 11270 / 300 = 37.6 A1 [4] Correct answer GCE AS/A LEVEL – May/June 2011 9709 63 4 (i) (a) P(final score is 12) = P(6, 6) = 1/36 B1 [1] Correct answer (b) P[(1,5) + (1,4) + (2,3) + (3,2) + (4,1)] M1 Considering P(1, 5) M1 Considering P[(1,4) + (2,3) + (3,2) + (4,1)] = 5/36 A1 [3] Correct answer (ii) P(A) = 1/6 P(B) = P[(1,5) + (2,4) + (3,3) + (4, 2) + (5,1)] = 5/36 B1 Any two of P(A), P(B) and P(C) correct P(C) = 1 – P(O, O) = 3/4 B1 Third probability correct P(A and B) = P(1 and 5) = 1/36 ≠ P(A) × P(B) M1 Numerical attempt to compare P(X and Y) P(A and C) = P[(2,5) + (4,5) + (6,5)] = 3/36 with P(X) × P(Y), must be three positive ≠ P(A) × P(C) probs P(B and C) = P[(2,4) + (4,2)] = 2/36 ≠ P(B) × P(C) None are independent. A1√ One correct comparison and conclusion, ft their probabilities A1 [5] Correct conclusion(s) following legitimate working 5 (i) z = ± 1.751 B1 Correct z 20 − µ ± = .1751 M1 Standardising no cc, no sqrt, must be a µ / 4 z-value µ = 13.9 A1 [3] Correct answer 10 − 13.91 (ii) P(X < 10) = P(z < ± ) M1 Standardising attempt with 10, their µ and 13.91 / 4 their µ/4, no cc, no sqrt = P(z < −1.124) M1 “Φ + Φ2 – 1”, ft their mean = 1− 0.8694 = 0.131 P(10 < X < 20) = 0.96 – 0.131 = 0.829 or 0.830 A1 [3] Correct answer (iii) µ = 250 × 0.96 = 240 B1 240 and 9.6 or sq rt 9.6 seen unsimplified σ2 = 250 × 0.96 × 0.04 = 9.6  234 5. − 240  P(≥ 235) = 1 − Φ  ±  M1 Standardising, with or without cc, must  6.9  have sq rt in denom M1 Continuity correction 234.5 or 235.5 only = Φ (1.775) M1 Correct region > 0.5, ft their mean = 0.962 A1 [5] Correct answer GCE AS/A LEVEL – May/June 2011 9709 63 6 (i) (0.75)n < 0.06 M1* Equation or inequality with 0.75n and 0.06 or 0.94 seen n > 9.78 M1dep* Attempt at solving by trial and error (can be implied) or using logarithms correctly n = 10 A1 [3] Correct answer (ii) E(X) = 14 × 0.75 or 10.5 M1 Evaluating binomial probability for an Try P(10) = 14C10(0.75)10(0.25)4 = 0.220 integer value directly above or below their mean P(11) = 14C11(0.75)11(0.25)3 = 0.240 M1 Evaluating the other binomial probability (mode is) 11 A1 [3] Correct answer OR M1 Evaluating binomial P(n) and P(n + 1) M1 Evaluating binomial P(10), P(11) and P(12) A1 Correct answer (iii) P(> 11) M1 A binomial term of the form = 14C12(0.75)12(0.25)2 + 14C13(0.75)13(0.25)1 14Cn pn(1 − p)14 – n seen, n ≠ 0 or 14 + (0.75)14 M1 Summing binomial P(12, 13, 14) or P(11, 12, 13, 14,) = 0.281 A1 Correct answer 0.280 – 0.282 P(3) = 5C3 (0.2811)3(0.7189)2 M1 A binomial term of the form 5C3p3(1 − p)2 seen, any p = 0.115 A1 [5] Correct answer

This question in 9709/63 May/June 2011

6 (a) Find the number of different ways in which the 12 letters of the word STRAWBERRIES can be arranged (i) if there are no restrictions, [2] (ii) if the 4 vowels A, E, E, I must all be together. [3] (b) (i) 4 astronauts are chosen from a certain number of candidates. If order of choosing is not taken into account, the number of ways the astronauts can be chosen is 3876. How many ways are there if order of choosing is taken into account? [2] (ii) 4 astronauts are chosen to go on a mission. Each of these astronauts can take 3 personal possessions with him. How many different ways can these 12 possessions be arranged in a row if each astronaut’s possessions are kept together? [2]

9 marks

Mark scheme: 12! M1 Dividing by 2! 3! 2! 6 (a) (i) = 19958400 (20,000,000) A1 [2] Correct answer !2!3!2 !4 !9 B1 4! seen multiplied (ii) × = 362880 B1 9! or 9 × 8! seen multiplied !2 !3!2 B1 [3] Correct final answer (b) (i) 3876 × 4! M1 Multiplying by 4! = 93024 A1 [2] Correct answer (ii) (3!)4 × 4! M1 3! or 6 or 4! seen = 31104 A1 [2] Correct final answer GCE AS/A LEVEL – October/November 2011 9709 61

This question in 9709/61 Oct/Nov 2011

2 Twelve coins are tossed and placed in a line. Each coin can show either a head or a tail. (i) Find the number of different arrangements of heads and tails which can be obtained. [2] (ii) Find the number of different arrangements which contain 7 heads and 5 tails. [1]

3 marks

Mark scheme: 12 seen2 (i) each in 2 ways = 212 M1 2 = 4096 A1 [2] Correct answer 12! (ii) B1 [1] 7!5! = 792

This question in 9709/62 Oct/Nov 2011

3 (a) Geoff wishes to plant 25 flowers in a flower-bed. He can choose from 15 different geraniums, 10 different roses and 8 different lilies. He wants to have at least 11 geraniums and also to have the same number of roses and lilies. Find the number of different selections of flowers he can make. [4] (b) Find the number of different ways in which the 9 letters of the word GREENGAGE can be arranged if exactly two of the Gs are next to each other. [3]

7 marks

Mark scheme: 3 (a) G R L 11 7 7 = 15C11 × 10C7 × 8C7 = 1310400 M1 Multiplying 3 combinations 13 6 6 = 15C13 × 10C6 × 8C6 = 617400 A1 One of 1310400, 617400, 14112 seen 15 5 5 = 15C15 × 10C5 × 8C5 = 14112 M1 Adding 3 options Total = 1941912 (1940000) A1 [4] Correct answer (b) e.g. * E * R * E (GG) N * A * E * gives 6 ways for G B1 7! / 3! Or 7!/3!3! seen oe 7! 6 or 8!/3! – 2 × 7!/3! B1 Multiplying by 6 (gaps) oe 3!× = 5040 ways. B1 [3] Correct final answer

This question in 9709/62 Oct/Nov 2011

4 Mary saves her digital images on her computer in three separate folders named ‘Family’, ‘Holiday’ and ‘Friends’. Her family folder contains 3 images, her holiday folder contains 4 images and her friends folder contains 8 images. All the images are different. (i) Find in how many ways she can arrange these 15 images in a row across her computer screen if she keeps the images from each folder together. [3] (ii) Find the number of different ways in which Mary can choose 6 of these images if there are 2 from each folder. [2] (iii) Find the number of different ways in which Mary can choose 6 of these images if there are at least 3 images from the friends folder and at least 1 image from each of the other two folders. [4]

9 marks

Mark scheme: 4 (i) 3! × 4! × 8! × 3! M1 Multiplying 3 factorials together M1 Multiplying by 3! = 34 836 480 (34 800 000) A1 [3] Correct answer (ii) 3C2×4C2×8C2 M1 Multiplying (only) 3 combinations together = 504 A1 [2] Correct answer (iii) Fr Fa H 3 1 2 = 8C3 × 3C1 × 4C2 = 1008 M1 Multiplying 3 combinations, only 3 2 1 = 8C3 × 3C2 × 4C1 = 672 M1 Summing 3 options 4 1 1 = 8C4 × 3C1 × 4C1 = 840 A1 3 correct combination answers total ways = 2520 A1 [4] Correct answer

This question in 9709/63 Oct/Nov 2011

5 An English examination consists of 8 questions in Part A and 3 questions in Part B. Candidates must choose 6 questions. The order in which questions are chosen does not matter. Find the number of ways in which the 6 questions can be chosen in each of the following cases. (i) There are no restrictions on which questions can be chosen. [1] (ii) Candidates must choose at least 4 questions from Part A. [3] (iii) Candidates must either choose both question 1 and question 2 in Part A, or choose neither of these questions. [3]

7 marks

Mark scheme: 5 (i) 11C6 = 462 B1 OR A3 B3 or A4 B2 or A5 B1 or A6 = 8C3 + 8C4 × 3C2 + 8C5 × 3C1 + 8C6 B1 = 56 + 210 + 168 + 28 = 462 [1] (ii) 8C4 × 3C2 + 8C5 × 3C1 + 8C6 M1 ∑ 2 or more two-factor terms, P or C any numbers = 210 + 168 + 28 B1 Any correct option unsimplified = 406 A1 [3] Correct answer (iii) 9C4 + 9C6 = 126 + 84 M1 Summing 9Cx + 9Cy can be mult by 2 no other terms B1 126 or 84 seen or unsimplified 9C4, 9C6 = 210 A1 Correct answer OR 1,2 in A tog with : A1B3 + A2B2 + A3B1 + M1 ∑ 5 or more 2-factor 6Px or 6Cx with 3Cx or A4B0 + 1,2 out of A : A3B3 + A4B2 + 3Px only (can be mult by 2) A5B1 + A6B0 = 6C1 + 6C2 × 3C2 + 6C3 × 3C1 + 6C4 + 6C3 × B1 3 or more correct unsimplified options 3C3 + 6C4 × 3C2 + 6C5 × 3C1 + 6C6 = 6 + 45 + 60 + 15 + 20 + 45 + 18 + 1 = 210 A1 Correct answer OR 462 – 9C5 – 9C5 M1 subt two 9Cx options from their (i) B1 9C5 seen oe if using this method = 210 A1 [3] Correct answer

This question in 9709/62 May/June 2012

5 (a) A team of 3 boys and 3 girls is to be chosen from a group of 12 boys and 9 girls to enter a competition. Tom and Henry are two of the boys in the group. Find the number of ways in which the team can be chosen if Tom and Henry are either both in the team or both not in the team. [3] (b) The back row of a cinema has 12 seats, all of which are empty. A group of 8 people, including Mary and Frances, sit in this row. Find the number of different ways they can sit in these 12 seats if (i) there are no restrictions, [1] (ii) Mary and Frances do not sit in seats which are next to each other, [3] (iii) all 8 people sit together with no empty seats between them. [3]

10 marks

Mark scheme: 5 (a) Boys in:10C1 × 9C3 = 840 ways M1 summing two 2-factor products, C or P Boys out: 10C3 × 9C3 = 10080 ways B1 Any correct option unsimplified Total = 10920 ways (10900) A1 Correct final answer [3] (b) (i) 12P8 = 19,958,400 B1 [1] or 20,000,000 (ii) together: 11P7 = 1663200 × 2 = 3326400 B1 11P7 seen Not tog: 19958400 – 3326400 M1 19958400 or their (i) – their together (must be >0) =16,632,000 (16,600,000) A1 [3] correct final answer OR M at end then not F in 10 × 10P6 × M1 summing options for M at end and M 2=3024000 ways not at end not at end in 10 × 9 × 10P6 = 13608000 B1 one correct option ways Total = 16,632,000 ways A1 correct final answer (iii) 8! × 5 = 201600 ways B1 8! seen mult by equivalent of integer ≥ 1 M1 Mult by 5 A1 Correct answer SR 8! × 5!=4838400 [3] B2 GCE AS/A LEVEL – October/November 2012 9709 62

This question in 9709/62 Oct/Nov 2012

6 Four families go to a theme park together. Mr and Mrs Lin take their 2 children. Mr O’Connor takes his 2 children. Mr and Mrs Ahmed take their 3 children. Mrs Burton takes her son. The 14 people all have to go through a turnstile one at a time to enter the theme park. (i) In how many different orders can the 14 people go through the turnstile if each family stays together? [3] (ii) In how many different orders can the 8 children and 6 adults go through the turnstile if no two adults go consecutively? [3] Once inside the theme park, the children go on the roller-coaster. Each roller-coaster car holds 3 people. (iii) In how many different ways can the 8 children be divided into two groups of 3 and one group of 2 to go on the roller-coaster? [3]

9 marks

Mark scheme: 6 (i) 4! × 3! × 5! × 2! × 4! = 829440 B1 4!, 3!, 5!, 2 seen multiplied 1, not in denominator B1 Mult by 4! B1 [3] Correct answer (ii) 8! × 9 × 8 × 7 × 6 × 5 × 4 B1 8! seen multiplied 1 B1 Mult by 9P6 = 2438553600 (2.44 × 109) B1 [3] Correct answer (iii) 8C3 × 5C3 × 2C2 B1 8C3 seen mult = 560 B1 5C3 seen mult B1 [3] Correct answer

This question in 9709/61 May/June 2013

7 There are 10 spaniels, 14 retrievers and 6 poodles at a dog show. 7 dogs are selected to go through to the final. (i) How many selections of 7 different dogs can be made if there must be at least 1 spaniel, at least 2 retrievers and at least 3 poodles? [4] 2 spaniels, 2 retrievers and 3 poodles go through to the final. They are placed in a line. (ii) How many different arrangements of these 7 dogs are there if the spaniels stand together and the retrievers stand together? [3] (iii) How many different arrangements of these 7 dogs are there if no poodle is next to another poodle? [3]

10 marks

Mark scheme: 10 7 7 × = M1 Mult two probabilities one containing x and

This question in 9709/63 May/June 2013

6 A shop has 7 different mountain bicycles, 5 different racing bicycles and 8 different ordinary bicycles on display. A cycling club selects 6 of these 20 bicycles to buy. (i) How many different selections can be made if there must be no more than 3 mountain bicycles and no more than 2 of each of the other types of bicycle? [4] The cycling club buys 3 mountain bicycles, 1 racing bicycle and 2 ordinary bicycles and parks them in a cycle rack, which has a row of 10 empty spaces. (ii) How many different arrangements are there in the cycle rack if the mountain bicycles are all together with no spaces between them, the ordinary bicycles are both together with no spaces between them and the spaces are all together? [3] (iii) How many different arrangements are there in the cycle rack if the ordinary bicycles are at each end of the bicycles and there are no spaces between any of the bicycles? [3]

10 marks

Mark scheme: 6 (i) M R O 3 1 2 = 7C3 × 5C1 × 8C2 = 4900 M1 Summing more than one 3term option involving combs (can be added) 3 2 1 = 7C3 × 5C2 × 8C1 = 2800 M1 Mult 3 combs only (indep) 2 2 2 = 7C2 × 5C2 × 8C2 = 5880 A1 1 option correct unsimplified Total = 13580 A1 4 Correct answer (ii) 4 groups in 4! ways M1 4! seen mult by something 3 mountain in 3! ways 2 ordinary in 2! ways M1 Mult by 3! for racing or 2! for ordinary 4! × 3! × 2 = 288 A1 3 Correct answer (iii) e.g. s O x x x x O s s s M1 2! or 4! seen mult Ordinary in 2! Rest of bikes in 4! M1 Mult by 5 (ssssb) Bikes and spaces 5 groups in 5 ways 2! × 4! × 5 = 240 A1 3 Correct answer GCE AS/A LEVEL – October/November 2013 9709 61

This question in 9709/61 Oct/Nov 2013

6 The 11 letters of the word REMEMBRANCE are arranged in a line. (i) Find the number of different arrangements if there are no restrictions. [1] (ii) Find the number of different arrangements which start and finish with the letter M. [2] (iii) Find the number of different arrangements which do not have all 4 vowels (E, E, A, E) next to each other. [3] 4 letters from the letters of the word REMEMBRANCE are chosen. (iv) Find the number of different selections which contain no Ms and no Rs and at least 2 Es. [3]

9 marks

Mark scheme: 6 (i) 1663200 B1 [1] (ii) M xxxxxxxxx M M1 9! or 9P9 seen !9 Number of ways = = 30240 A1 [2] Correct answer !2!3 (iii) 4 vowels together = 8! × 4/2!2! M1 8!/2!2! seen mult by something = 40320 M1 4 oe 4!/3! or 4C1 etc. seen mult by something 1663200 – 40320 = 1622880 B1 [3] Correct answer SC 7!/2!2! × 8P4 or 7! × 8P4/3! Or 7!/2!2! × 8P4/3! M1 (iv) Exactly 2 Es 4C2 = 6 M1 Summing 2 options Exactly 3 Es 4C1 = 4 B1 One option correct Total = 10 ways A1 [3] Correct answer OR 5C2 M2 M1 for k5C2 = 10 A1 Correct ans GCE AS/A LEVEL – October/November 2013 9709 62

This question in 9709/62 Oct/Nov 2013

6 (i) Find the number of different ways that the 9 letters of the word AGGREGATE can be arranged in a line if the first letter is R. [2] (ii) Find the number of different ways that the 9 letters of the word AGGREGATE can be arranged in a line if the 3 letters G are together, both letters A are together and both letters E are together. [2] (iii) The letters G, R and T are consonants and the letters A and E are vowels. Find the number of different ways that the 9 letters of the word AGGREGATE can be arranged in a line if consonants and vowels occur alternately. [3] (iv) Find the number of different selections of 4 letters of the word AGGREGATE which contain exactly 2 Gs or exactly 3 Gs. [3]

10 marks

Mark scheme: !8 6 (i) M1 8! Divided by at least one of 3!2!2! oe !2!2!3 = 1680 A1 2 Correct answer (ii) 5! M1 5! Seen (not added, may be divided/multipled) = 120 A1 2 Correct answer !4!5 (iii) B1 5! Or 4! Seen in sum or product in numerator !2!2!3 (denominator may by 1) k !4!5 M1 in a numerical expression !2!2!3 = 120 A1 3 Correct final answer (iv) GG with AA, AE, EE, RA, RE, RT, M1 Summing 2 options (could be lists) TA, TE, = 8 ways GGG with A, E, R, T = 4 ways A1 1 correct option Total = 12 ways A1 3 Correct answer GCE AS/A LEVEL – October/November 2013 9709 63

This question in 9709/63 Oct/Nov 2013

6 Find the number of different ways in which all 8 letters of the word TANZANIA can be arranged so that (i) all the letters A are together, [2] (ii) the first letter is a consonant (T, N, Z), the second letter is a vowel (A, I), the third letter is a consonant, the fourth letter is a vowel, and so on alternately. [3] 4 of the 8 letters of the word TANZANIA are selected. How many possible selections contain (iii) exactly 1 N and 1 A, [2] (iv) exactly 1 N? [3]

10 marks

Mark scheme: !6 B1 6! Seen alone 6 (i) = 360 !2 B1 2 Dividing by 2! only !4 !4 B1 4! seen mult (ii) × !2 !3 B1 Dividing by 2! or 3! (Mult by 4 implied B1B1) = 48 B1 3 Correct answer (iii) 1N and 1A: N A xx in 3C2 M1 3Cx or xC2 seen alone = 3 ways A1 2 Correct answer (iv) 0 A : Nxxx = 1 way M1 Finding ways with 0 or 2 or 3 As 2 As: NAAx in 3C1 = 3 ways M1 Summing 3 or 4 options 3 As: NAAA in 1 way Total = 8 ways A1 3 Correct answer GCE AS/A LEVEL – May/June 2014 9709 61

This question in 9709/61 May/June 2014

5 Find how many different numbers can be made from some or all of the digits of the number 1 345 789 if (i) all seven digits are used, the odd digits are all together and no digits are repeated, [2] (ii) the numbers made are even numbers between 3000 and 5000, and no digits are repeated, [3] (iii) the numbers made are multiples of 5 which are less than 1000, and digits can be repeated. [3]

8 marks

Mark scheme: 5 (i) 5! × 3! or 6! B1 5! or 3! or 6! oe seen mult or alone = 720 B1 2 Correct final answer (ii) 3**4, 3**8, 4**8 M1 considering at least 2 types of 4-figure options ending with 4 or 8 and starting with 3 or 4 B1 One option correct unsimplified can be implied = 5 × 4 + 5 × 4 + 5 × 4 = 60 A1 3 Correct final answer (iii) 5, *5, **5, M1 Appreciating that the number must end in 5 (can be implied) = 1 + 7 + 72 M1 summing numbers ending in 5 with at least 2 different numbers of digits = 57 A1 3 Correct final answer

This question in 9709/62 May/June 2014

7 Nine cards are numbered 1, 2, 2, 3, 3, 4, 6, 6, 6. (i) All nine cards are placed in a line, making a 9-digit number. Find how many different 9-digit numbers can be made in this way (a) if the even digits are all together, [4] (b) if the first and last digits are both odd. [3] (ii) Three of the nine cards are chosen and placed in a line, making a 3-digit number. Find how many different numbers can be made in this way (a) if there are no repeated digits, [2] (b) if the number is between 200 and 300. [2]

11 marks

Mark scheme: 7 (i) (a) 6! M1 Seen in a single term expression as numerator (×) 4! OR (×) 4 × 3 M1 Seen in a single term expression as numerator (denominator may be 1) ÷ 2!2!3! OR ÷ 2!3! M1 Seen in a single term expression as denominator Total 720 ways A1 4 Correct ans !7 !7 (i) (b) 1*******3 = = 420 B1 seen oe !2!3 !2!3 3*******1= 420 M1 Attempting to evaluate and sum at least 2 of 3*******3= 420 1***3, 3***1, 3***3 Total = 1260 ways A1 3 Correct ans (ii) (a) 5 × 4 × 3 = 60 ways (5P3) M1 5P3 or 5C3 ×3! (can be implied) A1 2 Correct ans (ii) (b) 2** in 212, 213, 214, 216, 221, 223, 224, 226, M1 Listing attempt starting with 2, at least 10 231, 232, 233, 234, 236, correct entries 241, 242, 243, 246 261, 262, 263, 264, 266 Total = 22 ways A1 2 Correct ans Alternative Methods: 3 × 4C1 + 2 × 5C1 M1 p × 4C1 + q × 5C1, oe p + q >2 OR OR 5P2 + 2C1 M1 5P2 seen OR OR 4P2 + 2 × 4P1 + 2C1 M1 Any 2 terms added

This question in 9709/63 May/June 2014

7 A committee of 6 people is to be chosen from 5 men and 8 women. In how many ways can this be done (i) if there are more women than men on the committee, [4] (ii) if the committee consists of 3 men and 3 women but two particular men refuse to be on the committee together? [3] One particular committee consists of 5 women and 1 man. (iii) In how many different ways can the committee members be arranged in a line if the man is not at either end? [3]

10 marks

Mark scheme: 7 (i) W(8) M(5) 4 2 = 8C4 × 5C2 = 700 M1 Mult 2 combs, 8Cx × 5Cy 5 1 = 8C5 × 5C1 = 280 M1 Summing 2 or 3 options 6 0 = 8C6 × 5C0 = 28 A1 2 correct options unsimplified Total = 1008 A1 4 Correct answer (ii) M1 and MMWWW = 3C2 × 8C3 = 168 M1 Summing 3 options M2 and MMWWW = 3C2 × 8C3 = 168 Neither and MMMWWW = 3C1 × 8C3 = B1 One correct option 56 Total = 392 A1 3 Correct answer M1 Subt 2 men together from no restrictions OR total, no restrictions = 5C3 ×8C3 = 560 B1 One correct of 560 or 168 M1M2 and MWWW = 3C1 × 8C3 = 168 A1 Correct answer 560 – 168 = 392 (iii) e.g. WWMWWW M1 5! Seen mult by integer > 1 = 5! (women) × 4 = 480 M1 Mult by 4 A1 3 Correct answer OR 6! – MWWWWW – WWWWWM M1 6! seen with a subtraction = 6! – 5! – 5! M1 5! or 2 × 5! Seen subtracted = 480 A1 Correct answer

This question in 9709/61 Oct/Nov 2014

6 (a) Seven fair dice each with faces marked 1, 2, 3, 4, 5, 6 are thrown and placed in a line. Find the number of possible arrangements where the sum of the numbers at each end of the line add up to 4. [3] (b) Find the number of ways in which 9 different computer games can be shared out between Wainah, Jingyi and Hebe so that each person receives an odd number of computer games. [6]

9 marks

Mark scheme: 6 (a) 1*****3 or 3*****1 or 2*****2 M1 Mult by 65 (for middle 5 dice outcomes) = 65 × 3 M1 Mult by 3 or summing 3 different combinations (for end dice outcomes) = 23328 A1 3 Correct answer accept 23 300 (b) W J H 1 1 7 = 9C1×8C1×1 = 72 M1 Multiplying 3 combinations (may be implied) 1 7 1 = 9C1×8C7×1 = 72 A1 1 unsimplified correct answer (72, 504, 1680, 216 or 3024) 7 1 1 = 9C7×2C1×1 = 72 1 3 5 = 9C1×8C3×1 = 504 mult by 3! A1 A 2nd unsimplified different correct answer 3 3 3 = 9C3×6C3×1 = 1680 M1 Summing options for 1,1.7 or 1,3,5 oe (mult by 3 or 3!) M1 Summing at least 2 different options of the 3 Total 4920 A1 6 Correct ans If no marks gained If games replaced M1M1M1 max available Listing all 10 different outcomes SCM1 If factorials used M0M1M1 max available

This question in 9709/63 Oct/Nov 2014

7 (a) Find how many different numbers can be made by arranging all nine digits of the number 223 677 888 if (i) there are no restrictions, [2] (ii) the number made is an even number. [4] (b) Sandra wishes to buy some applications (apps) for her smartphone but she only has enough money for 5 apps in total. There are 3 train apps, 6 social network apps and 14 games apps available. Sandra wants to have at least 1 of each type of app. Find the number of different possible selections of 5 apps that Sandra can choose. [5]

11 marks

Mark scheme: !9 7 (a) (i) B1 Dividing by 2!2!3! !3!2!2 2 = 15120 ways B1 [2] Correct answer !8 (ii) ********3 in = 1680 ways B1 Correct ways end in 3 !3!2!2 !8 ********7 in = 3360 ways B1 Correct ways end in 7 !3!2 Total even = 15120 – 1680 – 3360 M1 Finding odd and subt from 15120 or their (i) = 10080 ways A1 [4] Correct answer OR ********2 in 8!/2!3! = 3360 ways B1 One correct way end in even ********6 in 8!/2!2!3! = 1680 ways B1 correct way end in another even ********8 in 8!/2!2!2! = 5040ways M1 Summing 2 or 3 ways Total = 10080 ways A1 Correct answer OR “15120” ×6/9 = 10080 M2 Mult their (i) by 2/3 oe A2 Correct answer (b) T(3) S(6) G(14) 1 1 3 in 3×6×14C3 = 6552 M1 Mult 3 (combinations) together 1 3 1 in 3×6C3×14 = 840 assume 6 = 6C1etc 3 1 1 in 1×6×14 = 84 M1 Listing at least 4 different options 2 2 1 in 3C2×6C2×14 = 630 M1 Summing at least 4 different 2 1 2 in 3C2×6×14C2 = 1638 options 1 2 2 in 3×6C2×14C2 = 4095 B1 At least 3 correct numerical options Total ways = 13839 (13800) A1 [5] Correct answer

This question in 9709/61 May/June 2015

4 A group of 8 friends travels to the airport in two taxis, P and Q. Each taxi can take 4 passengers. (i) The 8 friends divide themselves into two groups of 4, one group for taxi P and one group for taxi Q, with Jon and Sarah travelling in the same taxi. Find the number of different ways in which this can be done. [3] Taxi P Taxi Q Back Front Back Front Each taxi can take 1 passenger in the front and 3 passengers in the back (see diagram). Mark sits in the front of taxi P and Jon and Sarah sit in the back of taxi P next to each other. (ii) Find the number of different seating arrangements that are now possible for the 8 friends. [4]

7 marks

Mark scheme: 4 (i) Two in same taxi: M1 6C4 or 6C2 oe seen anywhere 6C2 × 4C4 × 2 or 6C2 + 6C4 M1 'something' ×2 only or adding 2 equal terms = 30 A1 3 Correct final answer (ii) MJS in taxi M1 5P1, 5C1 or 5 seen anywhere (5C1×2×2)× 4P4 M1 Mult by 2 or 4 oe M1 Mult by 4P4 oe eg 4! or 4×3P3 or can be part of 5! = 480 A1 4 Correct final answer

This question in 9709/62 Oct/Nov 2015

5 (a) Find the number of different ways that the 13 letters of the word ACCOMMODATION can be arranged in a line if all the vowels (A, I, O) are next to each other. [3] (b) There are 7 Chinese, 6 European and 4 American students at an international conference. Four of the students are to be chosen to take part in a television broadcast. Find the number of different ways the students can be chosen if at least one Chinese and at least one European student are included. [5]

8 marks

Mark scheme: 5 (a) e.g. **(AAOOOI)***** B1 8! (8 × 7!) or 6! seen anywhere, either alone or in numerator) !8 !6 × = 604800 M1 Dividing by at least 3 of 2!2!2!3! (may be !2!2 !3!2 fractions added) A1 3 Correct answer (b) C(7) E(6) A(4) M1 Mult 3 appropriate combinations together 1 1 2 = 7 × 6 × 4C2 = 252 assume 6=6C1, 1=4C0 etc., ∑r=4, C&E both 1 2 1 = 7 × 6C2 × 4 = 420 present 1 3 0 = 7 × 6C3 × 1 = 140 2 1 1 = 7C2 × 6 × 4 = 504 A1 At least 3 correct unsimplified products 2 2 0 = 7C2 × 6C2 × 1 = 315 3 1 0 = 7C3 × 6 × 1 = 210 M1* Listing at least 4 different correct options DM1 Summing at least 4 outcomes, involving 3 combs or perms, ∑r=4 Total = 1841 A1 5 Correct answer SC if CE removed, M1 available for listing at least 4 different correct options for remaining 2. DM1 for 7C1×6C1×(sum of at least 4 outcomes)

This question in 9709/63 Oct/Nov 2015

6 Hannah chooses 5 singers from 15 applicants to appear in a concert. She lists the 5 singers in the order in which they will perform. (i) How many different lists can Hannah make? [2] Of the 15 applicants, 10 are female and 5 are male. (ii) Find the number of lists in which the first performer is male, the second is female, the third is male, the fourth is female and the fifth is male. [2] Hannah’s friend Ami would like the group of 5 performers to include more males than females. The order in which they perform is no longer relevant. (iii) Find the number of different selections of 5 performers with more males than females. [3] (iv) Two of the applicants are Mr and Mrs Blake. Find the number of different selections that include Mr and Mrs Blake and also fulfil Ami’s requirement. [3]

10 marks

Mark scheme: 6 (i) 15P5 M1 oe, can be implied Not 15C5 = 360360 A1 2 Correct answer (ii) 5 × 10 × 4 × 9 × 3 M1 Mult 5 numbers = 5400 A1 2 Correct answer (iii) M(5) F(10) 3 2 = 5C3 × 10C2 = 450 ways M1 Mult 2 combs, 5Cx × 10Cy 4 1 = 5C4 × 10C1 = 50 M1 Summing 2 or 3 two-factor options, 5 0 = 5C5 × 10C0 = 1 x + y =5 Total = 501 ways A1 3 Correct answer (iv) (Couple) M(4) F(9) ManWife + 3 0 = 4C3 × 9C0 = 4 M1 Mult 2 combs 4Cx and 9Cy ManWife + 2 1 = 4C2 × 9C1 = 54 M1 Summing both options x + y =3, gender correct Total = 58 A1 3 Correct answer

This question in 9709/62 Feb/March 2016

6 (a) (i) Find how many numbers there are between 100 and 999 in which all three digits are different. [3] (ii) Find how many of the numbers in part (i) are odd numbers greater than 700. [4] (b) A bunch of flowers consists of a mixture of roses, tulips and daffodils. Tom orders a bunch of 7 flowers from a shop to give to a friend. There must be at least 2 of each type of flower. The shop has 6 roses, 5 tulips and 4 daffodils, all different from each other. Find the number of different bunches of flowers that are possible. [4]

11 marks

Mark scheme: 6 (a) (i) 9 × 9 × 8 M1 M1 Logical listing attempt = 648 A1 [3] OR 900 – 28 × 9 = 648 (ii) (7….in 1 × 8 × 4 = 32 ways M1 Listing #s starting with 7 or 9 and ending odd 8 …in 1 × 8 × 5 = 40 M1 9… in 1 × 8 × 4 = 32 M1 Total 104 ways A1 [4] (b) R(6 ) T(5) D(4) 2 2 3 = 6C2×5C2×4C3 = 600 M1 Mult 3 combs, 6Cx×5Cy×4Cz 2 3 2 = 6C2×5C3×4C2 = 900 M1 Summing 2 or 3 three-factor outcomes 3 2 2 = 6C3×5C2×4C2 = 1200 can be perms, + instead of × A1 2 options correct unsimplified Total = 2700 A1 [4]

This question in 9709/61 May/June 2016

6 Find the number of ways all 9 letters of the word EVERGREEN can be arranged if (i) there are no restrictions, [1] (ii) the first letter is R and the last letter is G, [2] (iii) the Es are all together. [2] Three letters from the 9 letters of the word EVERGREEN are selected. (iv) Find the number of selections which contain no Es and exactly 1 R. [1] (v) Find the number of selections which contain no Es. [3]

9 marks

Mark scheme: 6 (i) 7560 ways B1 [1] 7! (ii) RxxxxxxxG in B1 7! alone seen in num or 4! alone in denom 4! 7!× 2 Must be in a fraction. gets full 4!× 2 marks B1 [2] = 210 ways 6! (iii) eg EEEExxxxx in B1 6! or 5! × 6 seen in numerator or on own 2! Can be 6! × k but not 6! ± k B1 [2] = 360 ways (iv) 1 R eg RVG or RVN or RGN = 3 B1 [1] (v) no Rs eg VGN or 3C3 ways = 1 M1 Summing at least 2 options for R 2 Rs eg RRV or 3C1 ways = 3 A1 Correct outcome for no Rs or 2 Rs – Total = 7 A1 [3] evaluated

This question in 9709/63 May/June 2016

5 (a) Find the number of different ways of arranging all nine letters of the word PINEAPPLE if no vowel (A, E, I) is next to another vowel. [4] (b) A certain country has a cricket squad of 16 people, consisting of 7 batsmen, 5 bowlers, 2 all- rounders and 2 wicket-keepers. The manager chooses a team of 11 players consisting of 5 batsmen, 4 bowlers, 1 all-rounder and 1 wicket-keeper. (i) Find the number of different teams the manager can choose. [2] (ii) Find the number of different teams the manager can choose if one particular batsman refuses to be in the team when one particular bowler is in the team. [3]

9 marks

Mark scheme: 5 (a) e.g. P*N*P*P*L M1 Mult by 5! in num 5! 6 P4 = × M1 Dividing by 3! or 2! 3! 2! M1 Mult by 6P4 oe = 3600 A1 [4] (b) (i) 7C5 × 5C4 × 2C1 × 2C1 M1 Mult 4 combs of which three are correct = 420 A1 [2] (ii) both in team M1 Evaluating both in team and subtracting from (i) 6C4 × 4C3 × 2 × 2 = 240 M1 240 seen can be unsimplified ft their 420, their 240 420 – 240 = 180 ways A1 OR Bat in bowl out + bowl in bat out + both out M1 summing 2 or 3 options not both in team = 6C4×4C3×2×2+6C5 × 4C3 × 2× 2+ 6C5 × 4C4 × 2 × 2 A1 2 or 3 options correct unsimplified = 60 + 96 + 24 = 180 ways A1 Correct ans from correct working OR Bat in bowl out + bat out M1 As above, or bowl in bat out + = 60 + 6C5× 5C4×2×2 = 60 + 120 = 180 ways A1 A1 [3] bowl out

This question in 9709/61 Oct/Nov 2016

6 Find the number of ways all 10 letters of the word COPENHAGEN can be arranged so that (i) the vowels (A, E, O) are together and the consonants (C, G, H, N, P) are together, [3] (ii) the Es are not next to each other. [4] Four letters are selected from the 10 letters of the word COPENHAGEN. (iii) Find the number of different selections if the four letters must contain the same number of Es and Ns with at least one of each. [5]

12 marks

Mark scheme: 6 (i) e.g. (OAEE)(CPNHGN) or cv M1 4!/2! or 6!/2! seen anywhere 4! 6! M1 All multiplied by 2 oe × × 2 = 8640 A1 [3] 2! 2! (ii) First Method Total ways = 10!/2!2! = 907200 B1 Total ways together correct EE together in 9!/2! ways = 181440 M1 EE together attempt alone EE not together = 907200 – 181440 M1 Considering total – EE together = 725760 A1 [4] OR Second Method C P N H G N O A in 8!/2! ways B1 8!/2! Seen Insert E in 9 ways M1 Interspersing an E, x n where n=7,8,9. Condone additional factors. Insert 2nd E in 8 ways, ÷ 2 M1 Mult by 9×8(÷2), 9C2 or 9P2 only oe Total = 8!/2!×9×8 ÷ 2 = 725760 A1 (iii) First Method EN** in 6C2 ways M1 6Cx or yC2 seen alone or mult by k > 1, x<6, y>2 M1 (1x1x) 6C2 seen strictly alone or added to their EENN only = 15 different ways A1 EENN in 1 way B1 Total 16 ways A1 [5] OR Second Method Listing with at least 8 different correct options M1 Value stated or implied by final answer Listing all correct options M1 Total = 15 different ways A1 correct value stated EENN in 1 way B1 Total 16 ways A1 Award 16 SRB2 if no method is present

This question in 9709/62 Oct/Nov 2016

1 A committee of 5 people is to be chosen from 4 men and 6 women. William is one of the 4 men and Mary is one of the 6 women. Find the number of different committees that can be chosen if William and Mary refuse to be on the committee together. [3]

3 marks

Mark scheme: 1 total ways 10C5 =252 M1 10C5 – … or 252– … MW together e.g. (MW)*** in 8C3 ways = 56 MW not together = 252 – 56 B1 252 and 56 seen, may be unsimplified = 196 ways A1 [3] OR 1 2 8C4+ 8C5 M1 2 nC4+ nC5 2 8C4 = 2x70=140; 8C5 = 56 B1 140 and 56 seen may be unsimplified 2 8C4+ 8C5 =196 A1 OR 2 2 9C5 –8C5 M1 2 9C5 – .. 2 9C5 = 2 × 126 = 252; 8C5 = 56 B1 252 and 56 seen, may be unsimplified 2 9C5 –8C5 =196 A1

This question in 9709/63 Oct/Nov 2016

3 Numbers are formed using some or all of the digits 4, 5, 6, 7 with no digit being used more than once. (i) Show that, using exactly 3 of the digits, there are 12 different odd numbers that can be formed. [3] (ii) Find how many odd numbers altogether can be formed. [3]

6 marks

Mark scheme: 3 (i) e.g. **5 in 3P2 ways = 6 M1 Recognising ends in 5 or 7, can be implied **7 in 3P2 = 6 M1 Summing ends in 5 + ends in 7 oe Total 12 AG A1 [3] Correct answer following legit working OR listing 457, 547, 467, 647, 567, 657, 475, 745 M1 Listing at least 5 different numbers ending in 465, 645, 675, 765 5 M1 Listing at least 5 different numbers ending in 7 Total 12 AG A1 (ii) 1 digit in 2 ways M1 Consider at least 3 options with different 2 digits in *5 or *7 = 3P1 × 2 = 6 number of digits. If no working, must be 3 or 4 from 2, 6, 12, 12 4 digits in ***5 or ***7 = 3P3 × 2 = 12 A1 One option correct from 1, 2 or 4 digits Total ways = 32 A1 [3]

This question in 9709/63 Oct/Nov 2016

5 (i) A plate of cakes holds 12 different cakes. Find the number of ways these cakes can be shared between Alex and James if each receives an odd number of cakes. [3] … … … … … … … … … … … … … … (ii) Another plate holds 7 cup cakes, each with a different colour icing, and 4 brownies, each of a different size. Find the number of different ways these 11 cakes can be arranged in a row if no brownie is next to another brownie. [3] … … … … … … … … … … … … … … … … (iii) A plate of biscuits holds 4 identical chocolate biscuits, 6 identical shortbread biscuits and 2 identical gingerbread biscuits. These biscuits are all placed in a row. Find how many different arrangements are possible if the chocolate biscuits are all kept together. [3] … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) M1 A1 Correct unsimplified answer (can be implied by final answer) = 2048 A1 Correct answer Total: 3 5(ii) 7! 8P4 B1 7! seen alone or multiplied only (cupcakes ordered) M1 multiplying by 8P4 o.e (placing brownies) = 8467200 A1 correct answer Total: 3 5(iii) 9! / (6! 2!) B1 9! oe seen alone or as numerator M1 dividing by at least one of 6!,2! (removing repeated shortbread or gingerbread biscuits) ignore 4! if present = 252 A1 correct answer Total: 3 × ×

This question in 9709/62 Feb/March 2017

7 (a) Eight children of different ages stand in a random order in a line. Find the number of different ways this can be done if none of the three youngest children stand next to each other. [3] … … … … … … … … … … (b) David chooses 5 chocolates from 6 different dark chocolates, 4 different white chocolates and 1 milk chocolate. He must choose at least one of each type. Find the number of different selections he can make. [4] … … … … … … … … … … … … (c) A password for Chelsea’s computer consists of 4 characters in a particular order. The characters are chosen from the following. ³ The 26 capital letters A to Z ³ The 9 digits 1 to 9 ³ The 5 symbols # ~ * ? ! The password must include at least one capital letter, at least one digit and at least one symbol. No character can be repeated. Find the number of different passwords that Chelsea can make. [4] … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) EITHER: e.g. xxxxx =5! for the other children (B1 Put y in 6 ways, then 5 then 4 for the youngest children B1 Mult by 6P3 OE Answer 5! × 6P3 = 14400 B1) Correct answer OR: total – 3 tog – 2 tog = 8! – 6!3! – 6! × 2 × 5 × 3 = 14400 (B1 8! – 6! × k ⩾1seen B1 6!3! or 6! × 2 × 5 × 3 seen subtracted B1) Correct answer Total: 3 7(b) D W M 2 2 1 = 6C2 × 4C2 × 1 = 90 B1 One correct unsimplified option 3 1 1 = 6C3 × 4 × 1 = 80 M1 Summing 2 or more 3-factor options which can contain perms or 3 factors added. The 1 can be implied 1 3 1 = 6 × 4C3 × 1 = 24 M1 Summing the correct 3 unsimplified outcomes only Total=194 ways A1 Total: 4 Question Answer Marks Guidance 7(c) C D S 2 1 1 = 26C2 × 9 × 5 × 4! = 351 000 M1 summing 2 or more options of the form (2 1 1), (1 2 1), (1 1 2), can have perms, can be added 1 2 1 = 26 × 9C2 × 5 × 4! = 112 320 M1 4 relevant products seen excluding 4! e.g. 26 × 9 × 8 × 5 or 26 × 9P2 × 5 for 2nd outcome, condone 26 × 9 × 5 × 37 as being relevant 1 1 2 = 26 × 9 × 5C2 × 4! = 56 160 M1 mult all terms by 4! or 4!/2! Total = 519 480 A1 Total: 4

This question in 9709/61 May/June 2017

6 (a) Find how many numbers between 3000 and 5000 can be formed from the digits 1, 2, 3, 4 and 5, (i) if digits are not repeated, [2] … … … … … … … … … … … (ii) if digits can be repeated and the number formed is odd. [3] … … … … … … … … … … … … (b) A box of 20 biscuits contains 4 different chocolate biscuits, 2 different oatmeal biscuits and 14 different ginger biscuits. 6 biscuits are selected from the box at random. (i) Find the number of different selections that include the 2 oatmeal biscuits. [2] … … … … … … … … … … (ii) Find the probability that fewer than 3 chocolate biscuits are selected. [4] … … … … … … … … … … … … …

11 marks

Mark scheme: 6(a)(i) M1 Total = 48 ways A1 Total: 2 6(a)(ii) 2 × 5 × 5 × 3 M1 M1 Seeing 52 mult; this mark is for correctly considering the middle two digits with replacement Mult by 6; this mark is for correctly considering the first and last digits = 150 ways A1 Totals: 3 Question Answer Marks Guidance 6(b)(i) OO**** in 18C4 ways M1 18Cx or the sum of five 2-factor products with n = 14 and 4, may be × by 2C2: 4C0 × 14C4 + 4C1 × 14C3 + 4C2 × 14C2 + 4C3 × 14C1 + 4C4 (× 14C0) = 3060 A1 Totals: 2 Question Answer Marks Guidance 6(b)(ii) Choc Not Choc 0 6= 1 × 16C6 = 8008 0.2066 1 5= 4C1 × 16C5 = 17472 0.4508 2 4= 4C2 × 16C4 = 10920 0.2817 OR Choc Oats Ginger 0 0 6 0 1 5 0 2 4 1 0 5 1 1 4 1 2 3 2 0 4 2 1 3 2 2 2 B1 The correct number of ways with one of 0, 1 or 2 chocs , unsimplified or any three correct number of ways of combining choc/oat/ginger, unsimplified Total = 36400 ways M1 sum the number of ways with 0, 1 and 2 chocs and two must be totally correct, unsimplified OR sum the nine combinations of choc, ginger, oats, six must be totally correct, unsimplified Probability = 36400/ 20 C6 M1 dividing by 20C6 (38760) oe = 0.939 (910/969) A1 Totals: 4 freq = fd × cw 10, 40, 120, 30

This question in 9709/63 May/June 2017

6 (a) A village hall has seats for 40 people, consisting of 8 rows with 5 seats in each row. Mary, Ahmad, Wayne, Elsie and John are the first to arrive in the village hall and no seats are taken before they arrive. (i) How many possible arrangements are there of seating Mary, Ahmad, Wayne, Elsie and John assuming there are no restrictions? [2] … … … … … … … (ii) How many possible arrangements are there of seating Mary, Ahmad, Wayne, Elsie and John if Mary and Ahmad sit together in the front row and the other three sit together in one of the other rows? [4] … … … … … … … … … … … … … … (b) In how many ways can a team of 4 people be chosen from 10 people if 2 of the people, Ross and Lionel, refuse to be in the team together? [4] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a)(i) M1 = 78 960 960 A1 2 6(a)(ii) not front row e.g. WEJ** in 3× 3! = 18 ways B1 3! seen mult by k⩾1 7 rows in 7 × 18= 126 ways B1 mult by 7 front row: e.g. *MA** in 4 × 2 = 8 ways M1 attempt at front row arrangements and multiplying by the 7 other rows arrangements, need not be correct Total 126×8 = 1008 A1 4 6(b) EITHER: e.g. *R** in 8C3 ways = 56 ways *L** in 8C3 = 56 ways (M1 Considering either R or L only in team **** in 8C4 = 70 ways M1* Considering neither in team DM1 summing 3 scenarios Total 182 ways A1) OR1: No restrictions 10C4 = 210 ways (M1 10C4 – , Considering no restrictions with subtraction *RL* = 8C2 = 28 M1* Considering both in team 210 – 28 DM1 subt = 182 ways A1) Question Answer Marks Guidance 6(b) OR2: R out in 9C4 = 126 ways L out in 9C4 = 126 ways (M1 Considering either R out or L out Both out in 8C4 = 70 M1* Considering both out DM1 Summing 2 scenarios and subtracting 1 scenario 126 + 126 – 70 = 182 ways. A1) 4

This question in 9709/61 Oct/Nov 2017

6 A car park has spaces for 18 cars, arranged in a line. On one day there are 5 cars, of different makes, parked in randomly chosen positions and 13 empty spaces. (i) Find the number of possible arrangements of the 5 cars in the car park. [2] … … … … … (ii) Find the probability that the 5 cars are not all next to each other. [5] … … … … … … … … … … … … … … … … … On another day, 12 cars of different makes are parked in the car park. 5 of these cars are red, 4 are white and 3 are black. Elizabeth selects 3 of these cars. (iii) Find the number of selections Elizabeth can make that include cars of at least 2 different colours. [5] … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(i) M1 = 1 028 160 A1 2 Question Answer Marks Guidance 6(ii) EITHER: e.g. ***(CCCCC)********** in 5!×14 ways (B1 5! OE mult by k ⩾ 1, considering the arrangements of cars next to each other = 1680 B1 Mult by 14 OE, (or 14 on its own) considering positions within the line P (next to each other) = 1680/1 028 160 M1 Dividing by (i) for probability P(not next to each other) = 1 – 1680/1 028 160 M1 Subtracting prob from 1 (or their ‘5! 14 × ’ from (i) ) = 0.998 611 612       OE A1) OR1: 5! 14! 18! × = 0.001634 (B1 5! OE mult by k ⩾ 1 (on its own or in numerator of fraction) considering the arrangements of cars next to each other B1 Multiply by 14!, (or 14! on its own) considering all ways of arranging spaces with 5 cars together M1 Dividing by 18!, total number of ways of arranging spaces 1 – 0.001634 M1 Subtracting prob from 1 (or ‘5! × 14!’ from 18!) = 0.998(366) A1) OR2: 4 together – 2 5! 14 12 21 840 × × = C 3, 1, 1 – 3 5! 14 11 131040 × × = C 3, 2 – 2 5! 14 12 21840 × × = C 2,2,1 – 3 5! 14 11 131040 × × = C 2,1,1,1 – 4 5! 14 10 480 480 × × = C 1,1,1,1,1 – 5! 14 9 1 4 5 240 240 × = C or P (M1 Listing the six correct scenarios (only): 4 together; 3 together and 2 separate; 3 together and 2 together; two sets of 2 together and 1 separate; 2 together and 3 separate; 5 separate. M1 Summing total of the six scenarios, at least 2 correct unsimplified Question Answer Marks Guidance Total = 1 026 480 A1 Total of 1 026 480 M1 Dividing their 1 026 480 by their 6(i) 1 026 480 ( ) 1028160 0.998 366 ÷ = A1) 5 Question Answer Marks Guidance 6(iii) R(5) W(4) B(3) Scenarios No. of ways 1 1 1 = 5 × 4 × 3 = 60 0 1 2 = 4 × 3C2 = 12 0 2 1 = 4C2 × 3 = 18 1 0 2 = 5 × 3C2 = 15 2 0 1 = 5C2 × 3 = 30 1 2 0 = 5 × 4C2 = 30 2 1 0 = 5C2 × 4 = 40 B1 5 1 4 1 3 1 × × C C C or better seen i.e. no. of ways with 3 different colours M1 Any of 5C2 or 4C2 or 3C2 seen multiplied by k > 1 (can be implied) A1 2 correct unsimplified ‘no. of ways’ other than 5C1 × 4C1 × 3C1 M1 Summing no more than 7 scenario totals containing at least 6 correct scenarios Total = 205 A1 OR 12C3 – M1 Seeing ‘12C3 –’, considering all selections of 3 cars – 5C3 M1 Subt 5C3 OE, removing only red selections – 4C3 M1 Subt 4C3 OE, removing only white selections – 3C3 M1 Subt 3C3 OE, removing only black selections = 205 A1 Correct answer 5

This question in 9709/63 Oct/Nov 2017

2 A selection of 3 letters from the 8 letters of the word COLLIDER is made. (i) How many different selections of 3 letters can be made if there is exactly one L? [1] … … … … … … (ii) How many different selections of 3 letters can be made if there are no restrictions? [3] … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2(i) B1 1 2(ii) No L: 6C3 = 20 (1 L: 6C2 = 15) M1 Either 0L or 2L correct unsimplified 2 L: 6C1 = 6 M1 Summing the 3 correct scenarios Total = 41 A1 3

This question in 9709/62 Feb/March 2018

6 The digits 1, 3, 5, 6, 6, 6, 8 can be arranged to form many different 7-digit numbers. (i) How many of the 7-digit numbers have all the even digits together and all the odd digits together? [3] … … … … … … … … … … … (ii) How many of the 7-digit numbers are even? [3] … … … … … … … … … … …

6 marks

Mark scheme: 6(i) 3! × 4! 3! × 2 M1 4!/3! oe seen multiplied by integer > 1, no addition = 48 A1 3 6(ii) EITHER: Even = Total number of arrangements – Odd numbers = 7!/3! – 3× 6 5 4 3 2 1 3! × × × × × = (7!/3! – 6!/2!) = 840 −360 B1 7!/3! – B1 6!/2! OE = 480 B1 OR: No of arrangements ending in 8: 6! 3! B1 No. ending in 8 or no. ending in 6 correct unsimplified No ending in 6: 6!/2! B1 Both correct and added unsimplified Total: 6! 6!/ 2 120 360 480 3! + = + = B1 3

This question in 9709/62 Feb/March 2018

7 Find the number of different ways in which all 9 letters of the word MINCEMEAT can be arranged in each of the following cases. (i) There are no restrictions. [1] … … … … … … … (ii) No vowel (A, E, I are vowels) is next to another vowel. [4] … … … … … … … … … … … … … … … 5 of the 9 letters of the word MINCEMEAT are selected. (iii) Find the number of possible selections which contain exactly 1 M and exactly 1 E. [2] … … … … … … (iv) Find the number of possible selections which contain at least 1 M and at least 1 E. [3] … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(i) 9! 2!2! = 90720 B1 1 7(ii) Method 1 * * * * * A B1 5! seen multiplied (arrangement of consonants allowing repeats) No. arrangements of consonants × ways of inserting vowels = B1 6P4 oe (i.e. 6 × 5 × 4 × 3, 6C4 × 4!) seen mult (allowing repeats) no extra terms 5 2 ! ! × 6 4P 2! B1 Dividing by at least one 2! (removing at least one set of repeats) Answer 6 4P 5 2! 2 × = 10 800 B1 Correct final answer 4 7(iii) 5C3 = 10 M1 5Cx or 5Px seen alone, x = 2 or 3 A1 Correct final answer not from 5C2 2 Question Answer Marks Guidance 7(iv) Method 1 Considering separate groups M1 Considering two scenarios of MME or EEM or MMEE with attempt, may be probs or perms MME** = 5C2 = 10 MEE** = 5C2 = 10 MMEE* = 5C1 = 5 M1 Summing three appropriate scenarios from the four need 5Cx seen in all of them ME*** = 5C3 = 10 see (iii) Total = 35 A1 Correct final answer Method 2 Considering criteria are met if ME are chosen M1 7Cx only seen, no other terms M1 xC3 only seen, no other terms ME *** = 7C3 = 35 A1 Correct final answer 3

This question in 9709/61 May/June 2018

6 (a) Find the number of ways in which all 9 letters of the word AUSTRALIA can be arranged in each of the following cases. (i) All the vowels (A, I, U are vowels) are together. [3] … … … … … … … … … … … (ii) The letter T is in the central position and each end position is occupied by one of the other consonants (R, S, L). [3] … … … … … … … … … … … (b) Donna has 2 necklaces, 8 rings and 4 bracelets, all different. She chooses 4 pieces of jewellery. How many possible selections can she make if she chooses at least 1 necklace and at least 1 bracelet? [4] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a)(i) (AAAIU) * * * * Arrangements of vowels/repeats × arrangements of (consonants & vowel group) = M1 5!× 5! 3! M1 ! 3 m (m is an integer, m ⩾ 1) Both Ms can only be awarded if expression is fully correct = 2400 A1 Correct answer 3 6(a)(ii) E.g. R * * * T * * * L . Arrangements of consonants RL, RS, SL = 3P2 = 6 Arrangements of remaining letters = 6! 3! = 120 M1 k × 6! 3! or k × 3P2 or k × 3C2 or k × 3! or k × 3 × 2 (k is an integer, k ⩾ 1), no irrelevant addition Total 120 × 6 M1 Correct unsimplified expression or 6! 3! × 3C2 = 720 ways A1 Correct answer 3 Question Answer Marks Guidance 6(b) Method 1 N(2) R(8) Br(4) 1 2 1 = 2×8C2×4 = 224 M1 Multiply 3 combinations, 2Cx×8Cy×4Cz. Accept 2C1 = 2 etc. 2 1 1 = 1×8C1×4 = 32 1 1 2 = 2×8×4C2 = 96 A1 3 or more options correct unsimplified 2 0 2 = 1×1×4C2 = 6 1 0 3 = 2×1×4 = 8 M1 Summing their values of 4 or 5 legitimate scenarios (no extra scenarios) Total = 366 ways A1 Correct answer Method 2 14C4 – (2N2R or 1N3R or 4R or 3R1B or 2R2B or 1R3B or 4B) M1 ‘14C4 – k’ seen, k an integer from an expression containing 8Cx 1001 – (1×8C2 + 2×8C3 + 8C4 + 8C3×4 + 8C2×4C2 + 8×4 +1) A1 4 or more ‘subtraction’ options correct unsimplified, may be in a list 1001 – (28 + 112 + 70 + 224 + 168 + 32 + 1) M1 Their 14C4 – [their values of 6 or more legitimate scenarios] (no extra scenarios, condone omission of final bracket) = 366 A1 Correct answer 4

This question in 9709/62 May/June 2018

7 Find the number of ways the 9 letters of the word SEVENTEEN can be arranged in each of the following cases. (i) One of the letter Es is in the centre with 4 letters on either side. [2] … … … … … … … (ii) No E is next to another E. [3] … … … … … … … … … … … … … … … 5 letters are chosen from the 9 letters of the word SEVENTEEN. (iii) Find the number of possible selections which contain exactly 2 Es and exactly 2 Ns. [1] … … … … (iv) Find the number of possible selections which contain at least 2 Es. [4] … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(i) ****E**** Other letters arranged in 8! 2!3! = 3360 ways M1 A1 Correct final answer www OR 8× 7× 6×5× 4× 4×3× 2×1 4!2! = 3360 ways M1 Correct numerator (161 280) A1 Correct final answer www Total: 2 7(ii) * * * * * Arrangements other letters × ways Es inserted = 5! 2! × 6 4 C 6 4 5! 2! 4! P   ×     M1 k mult by 6 4 C or 6 4P oe (ways to insert Es ignoring repeats), k can = 1 or k mult by 5! 2! M1 Correct unsimplified expression or 5! 2! × 6 4P = 900 ways A1 Correct answer OR Total no of ways – no of ways with Es touching 9!/(4! × 2!) – … or 7 560 – … 6! 2! + 6 2 5 2 P ! × ! + 6 2P 5! 2! 2! × + 6 3P 5! 2! 2! × = 360 + 1800 + 900 + 3600 = 6660 M1 7560 unsimplified – k M1 Attempting to find four ways of Es touching (4 Es, 3Es and a single, 2 lots of 2 Es, 2 Es and 2 singles) 7 560 – 6 660 = 900 A1 Correct answer Question Answer Marks Guidance 7(ii) OR Adding the number of ways with the first E in the 1st (E1), 2nd (E2) or 3rd (E3) position. 5! 2! (E1 + E2 + E3) where E1 = 10, E2 = 4, E3 = 1 5! 2! (E1 + E2 + E3) M1 For any values for E1, E2 and E3 M1 For any two correct values of E1, E2 and E3 600 + 240 + 60 = 900 A1 Correct answer Total: 3 7(iii) EENN* in 3 ways B1 Numerical value must be stated Total: 1 Question Answer Marks Guidance 7(iv) EE *** with no N: 1 way EEN** 3C2 or listing 3 ways EENN* 3 ways from (iii) M1 Identifying the three different scenarios of EE, EEE or EEEE A1 Total no of ways with two Es (7 or 3 + 3 + 1) EEE** with no N: 3 ways EEEN* 3 ways EEENN 1 way A1 Total no. of ways with 3 Es (7) EEEE* no N 3 ways EEEEN 1 way Total 18 ways A1 Correct answer stated Method List containing ways with 2Es, 3Es and 4Es List containing at least 8 correct different ways List of all 18 correct ways Total 18 M1 At least 1 option listed for each of EE^^^, EEE^^, EEEE^ A1 Ignore repeated options A1 Ignore repeated/incorrect options A1 Correct answer stated Total: 4

This question in 9709/63 May/June 2018

1 9 people are to be divided into a group of 4, a group of 3 and a group of 2. In how many different ways can this be done? [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 B1 =126 × 10 × 1 B1 5 or 7C3 or 6 or 7C4 or 6 or 5C2 times an integer (2nd group) =1260 B1 Correct answer 3

This question in 9709/61 Oct/Nov 2018

3 In an orchestra, there are 11 violinists, 5 cellists and 4 double bass players. A small group of 6 musicians is to be selected from these 20. (i) How many different selections of 6 musicians can be made if there must be at least 4 violinists, at least 1 cellist and no more than 1 double bass player? [4] … … … … … … … … … … … … … … … … … … … … … … … The small group that is selected contains 4 violinists, 1 cellist and 1 double bass player. They sit in a line to perform a concert. (ii) How many different arrangements are there of these 6 musicians if the violinists must sit together? [3] … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(i) Scenarios are: 4V + 1C + 1DB: 11C4 × 5C1 × 4C1 M1 4V + 2C: 11C4 × 5C2 5V + 1C: 11C5 × 5C1 B1 2 correct unsimplified options 6600 + 3300 + 2310 M1 Add 2 or 3 correct scenarios only = 12210 A1 Correct answer 4 3(ii) 4! × 3! M1 k multiplied by 3! or 4!, k an integer ⩾ 1 A1 Correct unsimplified expression = 144 A1 Correct answer 3

This question in 9709/61 Oct/Nov 2018

1 (i) How many different arrangements are there of the 11 letters in the word MISSISSIPPI? [2] … … … … … … … … … (ii) Two letters are chosen at random from the 11 letters in the word MISSISSIPPI. Find the probability that these two letters are the same. [3] … … … … … … … … … … … … … …

5 marks

Mark scheme: 1(i) 11! 4!4!2! M1 11! 11! 4! 2! or k k × × , k a positive integer = 34650 A1 Correct final answer 2 1(ii) Method 1 P(SS) = 4 3 12 11 10 110 × = (= 0.10911) B1 One of P(SS), P(PP) or P(II) correct, allow unsimplified P(PP) = 2 1 2 11 10 110 × = (= 0.01818) P(II) = 4 3 12 11 10 110 × = (= 0.10911) 4 3 11 10 × M1 Sum of probabilities from 3 appropriate identifiable scenarios (either by labelling or of form 4 2 4 11 11 11 a c a b b b × + × + × where a = 4 or 3, b = 11 or 10, c = 2 or 1) Total = 26 13 110 55 = oe (0.236) A1 Correct final answer Method 2 Total number of selections = 11C2 = 55 Selections with 2 Ps = 1 B1 Seen as the denominator of fraction (no extra terms) allow unsimplified Selections with 2 Ss = 4C2 = 6 Selections with 2 Is = 4C2 = 6, M1 Sum of 3 appropriate identifiable scenarios (either by labelling or values, condone use of permutations. May be implied by 2,12,12) Total selections with 2 letters the same = 13 Probability of 2 letters the same = 13 55 oe (0.236) A1 Correct final answer, without use of permutations 3

This question in 9709/62 Oct/Nov 2018

4 (i) Find the number of different ways that 5 boys and 6 girls can stand in a row if all the boys stand together and all the girls stand together. [3] … … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the number of different ways that 5 boys and 6 girls can stand in a row if no boy stands next to another boy. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(i) 5! × 6! ×2 B1 n × 5! × 6! (n integer, n ⩾ 1), no inappropriate addition = 172800 B1 Correct final answer, isw rounding (www scores B3) All marks based on their final answer 3 Question Answer Marks Guidance 4(ii) … G … G … G … G … G … G … No. ways girls placed × No. ways boys placed in gaps = M1 k × 6! or k × 7P5 (k is an integer, k ⩾ 1) no inappropriate add. (7P5 ≡ 7 × 6 × 5 × 4 × 3 or 7C5 × 5!) 6! × 7P5 M1 Correct unsimplified expression = 1814400 A1 Correct exact final answer (ignore subsequent rounding) 3

This question in 9709/62 Oct/Nov 2018

1 A group consists of 5 men and 2 women. Find the number of different ways that the group can stand in a line if the women are not next to each other. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 Method 1 ... M … M … M … M … M … M1 k × 5! (120) or k × 6P2 (30), k is an integer ⩾ 1, No. ways men placed × No. ways women placed in gaps = 5! × 6P2 M1 Correct unsimplified expression = 3600 A1 Correct answer Method 2 Number with women together = 6! × 2 (1440) Total number of arrangements = 7! (5040) M1 6! × 2 or 7! – k seen, k is an integer ⩾ 1 Number with women not together = 7! – 6! × 2 M1 Correct unsimplified expression = 3600 A1 Correct answer 3

This question in 9709/63 Oct/Nov 2018

4 Out of a class of 8 boys and 4 girls, a group of 7 people is chosen at random. (i) Find the probability that the group of 7 includes one particular boy. [3] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the probability that the group of 7 includes at least 2 girls. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) B1 Seen as denominator of fraction Selections with boy included = 11C6 or 12C7 – 11C7 = 462 M1 Correct unsimplified expression for selections with boy included seen as numerator of fraction Probability = 462/792 = 7/12 (0.583) A1 Correct answer OR prob of boy not included = 11/12 × 10/11 × …. × 5/6= 5/12 B1 Correct unsimplified prob 1 – 5/12 M1 Subtracting prob from 1 = 7/12 A1 Correct answer 3 Question Answer Marks Guidance 4(ii) Method 1 Scenarios are: 2G + 5B: 4C2 × 8C5 = 336 B1 One unsimplified product correct 3G + 4B: 4C3 × 8C4 = 280 4G + 3B: 4C4 × 8C3 = 56 M1 No of selections (products of n C r and n P r ) added for 2, 3 and 4 girls with no of girls and no of boys summing to 7 Total = 672 A1 Correct total Probability = 672/792 (28/33) (0.848) A1ft Correct answer – ‘total’/( ‘total no of selections’ from i) Method 2 0G + 7B 4C0 × 8C7 = 8 B1 One unsimplified no of selections correct 1G + 6B 4C1 × 8C6 = 112 Total = 8 + 112 = 120 M1 No of selections (products of n C r and n P r ) added for 0 and 1 girls with no of girls and no of boys summing to 7 (12C7 – 120)/792 or 1 – 120/792 A1 792 – 120 = 672 or 1 – 120/792 Probability = 672/792 (28/33) (0.848) A1ft ‘672’ over ‘792’ from i Method 3 (probability) 1 – P(0) – P(1) = 1 – (8/12 × 7/11 × …… × 2/6) – (8/12 × … × 3/7 × 4/6 × 7) B1 One correct unsimplified prob for 0 or 1 = 1 – 1/99 –14/99 M1 Subtracting ‘P(0)’ and ‘P(1)’ (using products of 7 fractions with denominators from 12 to 6) from 1 A1 Both probs correct unsimplified = 84/99 = 28/33 A1ft 1 – ‘P(0)’ – ‘P(1)’ Question Answer Marks Guidance 4(ii) Method 4 (probability) P(2) + P(3) + P(4) = B1 One correct unsimplified prob for 2, 3 or 4 42/99 + 35/99 + 7/99 M1 Adding ‘P(2)’, ‘P(3)’ and P(4)’ (using products of 7 fractions with denominators from 12 to 6) A1 Three probs correct unsimplified = 84/99 = 28/33 A1ft ‘P(2)’+ ‘P(3)’ + ‘P(4)’ 4

This question in 9709/63 Oct/Nov 2018

7 Find the number of different arrangements that can be made of all 9 letters in the word CAMERAMAN in each of the following cases. (i) There are no restrictions. [2] … … … … … (ii) The As occupy the 1st, 5th and 9th positions. [1] … … … … (iii) There is exactly one letter between the Ms. [4] … … … … … … … … … … … … Three letters are selected from the 9 letters of the word CAMERAMAN. (iv) Find the number of different selections if the three letters include exactly one M and exactly one A. [1] … … … … … (v) Find the number of different selections if the three letters include at least one M. [3] … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(i) 9! 2!3! = 30240 A1 Exact value, final answer 2 7(ii) A ^ ^ ^ A ^ ^ ^ A Arrangements = 6! 360 2! = B1 Final answer 1 7(iii) M ^ M ^ ^ ^ ^ ^ ^ = 7! 7 3!× M1 7! in numerator, (considering letters not M) M1 Division by 3! only (removing repeated As) M1 Multiply by 7 (positions of M-M) = 5880 A1 Exact value, final answer Method 2 (choosing letter between Ms) 6! 6! 1 7 4 7 2! 3! × × + × × M1 6! in sum of 2 expressions a6! + b6! M1 Multiply by 7 in both expressions (positions of M-M) = 2520 + 3360 M1 2! 3! + c d seen (removing repeated As) = 5880 A1 Exact value Question Answer Marks Guidance 7(iii) Method 3 (MAM) ^ ^ ^ ^ ^ ^ = 7!/2! = 2520 M1 7! in numerator (considering 6 letters + block) (MA’M) ^ ^ ^ ^ ^ ^ = 7!/3! × 4 = 840 × 4 = 3360 M1 Division by 2! and 3! seen in different terms Total = 2520 + 3360 M1 Summing 5 correct scenarios only = 5880 A1 Exact value 4 7(iv) M A ^ = 4C1 = 4 B1 Final answer 1 7(v) M ^ ^ : 4C2 = 6 M M ^ : 4C1 = 4 M1 Either option M M ^ or M ^ ^ correct, accept unsimplified M M A : = 1 M A A : = 1 (M A _ :4C1 = 4) M1 Add 4 or 5 correct scenarios only Total = 16 A1 Value must be clearly stated Method 2 M M ^ = 5C1 = 5 M1 Either option M M ^ or M ^ ^ correct, accept unsimplified M ^ ^ = 5C2 = 10 M1 Adding 2 or 3 correct scenarios only M A A = = 1 Total = 16 A1 Value must be clearly stated 3

This question in 9709/62 Feb/March 2019

8 Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him. (i) In how many different ways can Freddie choose 4 toys? [1] … … … (ii) How many of these choices will include both his favourite car and his favourite bus? [2] … … … … … Freddie arranges these 9 toys in a line. (iii) Find the number of possible arrangements if the buses are all next to each other. [3] … … … … … … … … … … … … (iv) Find the number of possible arrangements if there is a car at each end of the line and no buses are next to each other. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(i) B1 1 8(ii) 7C2 B1 7Cx or yC2 (implied by correct answer) or 7Px or 7Py, seen alone = 21 B1 correct answer 2 Question Answer Marks Guidance 8(iii) _ C1 (B1 B2 B3 ) C2 _ C3 _ C4 _ C5 _ C6 B1 3! or 6! seen alone or multiplied by k > 1 need not be an integer 3! × 6! × 7 B1 3! and 6! seen multiplied by k > 1, integer, no division = 30240 B1 Exact value Alternative method for question 8(iii) C1 (B1 B2 B3 ) C2 C3 C4 C5 C6 B1 3! or 7! seen alone or multiplied by k > 1 need not be an integer 3! × 7! B1 3! and 7! seen multiplied by k > or = 1, no division = 30240 B1 Exact value 3 8(iv) C1 _ C2 _ C3 _ C4 _ C5 _C6 B1 6! or 4! X 6P2 seen alone or multiplied by k > 1, no division (arrangements of cars) 6! × 5P3 or 6! × 5 × 4 × 3 or 6! x 3! x10 B1 Multiply by 5P3 oe i.e. putting Bs in between 4 of the Cs OR multiply by 3! x n where n = 7, 8, 9, 10 (number of options) = 43200 B1 Correct answer 3

This question in 9709/61 May/June 2019

7 (a) A group of 6 teenagers go boating. There are three boats available. One boat has room for 3 people, one has room for 2 people and one has room for 1 person. Find the number of different ways the group of 6 teenagers can be divided between the three boats. [3] … … … … … … … … … … … … … … … (b) Find the number of different 7-digit numbers which can be formed from the seven digits 2, 2, 3, 7, 7, 7, 8 in each of the following cases. (i) The odd digits are together and the even digits are together. [3] … … … … … … … … … … … … … (ii) The 2s are not together. [4] … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) M1 condone use of permutations, = 20 × 3 A1 Any correct method seen no addition/additional scenarios = 60 A1 Correct answer Alternative method for question 7(a) 6 6 3 2 1 3 2 1 P 6! 3! 2! P P P = × × × M1 6P6 / (nPn x k) with 3 ⩾ n > 1and 6 ⩾ k an integer ⩾ 1, not 6!/1 A1 Correct method with no additional terms = 60 A1 Correct answer 3 7(b)(i) 4! 3! 2 3! 2! × × M1 A single expression with either 4!/3! × k or 3!/2! × k, k a positive integer seen oe (condone 2 identical expressions being added) M1 Correctly multiplying their single expression by 2 or 2 identical expressions being added. = 24 A1 Correct answer 3 Question Answer Marks Guidance 7(b)(ii) Total no of arrangements = 7! 2!3! = 420 (A) B1 Accept unsimplified No with 2s together = 6! 3! = 120 (B) B1 Accept unsimplified With 2s not together: their (A) – their (B) M1 Subtraction indicated, possibly by their answer, no additional terms present = 300 ways A1 Exact value www Alternative method for question 7(b)(ii) 3 _ 7 _ 7 _ 7 _ 8 _ 5! 6 5 3! 2 × × B1 k x 5! in numerator, k a positive integer B1 m x 3! In denominator, m a positive integer M1 Their 5!/3! multiplied by 6C2 only (no additional terms) = 300 ways A1 Exact value www 4

This question in 9709/62 May/June 2019

3 Mr and Mrs Keene and their 5 children all go to watch a football match, together with their friends Mr and Mrs Uzuma and their 2 children. Find the number of ways in which all 11 people can line up at the entrance in each of the following cases. (i) Mr Keene stands at one end of the line and Mr Uzuma stands at the other end. [2] … … … … … … … … … … (ii) The 5 Keene children all stand together and the Uzuma children both stand together. [3] … … … … … … … … … … … …

5 marks

Mark scheme: 3(i) 9! × 2 = 725760 B1 Exact value 2 3(ii) Eg (K1K2K3K4K5) A A A (U1U2) A = 5! × 2! × 6! B1 2! or 5! seen mult by k > 1, no addition (arranging Us or Ks) B1 6! Seen mult by k > 1, no addition (arranging AAAAKU) = 172800 B1 Exact value 3

This question in 9709/63 May/June 2019

4 (i) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if there must be at least twice as many men as there are women on the committee. [3] … … … … … … … … … … … … (ii) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if 2 particular men refuse to be on the committee together. [3] … … … … … … … … … … …

6 marks

Mark scheme: 4(i) M(8) W(4) 4 2 in 8C4 × 4C2 = 420 ways 5 1 in 8C5 × 4C1 = 224 ways 6 0 in 8C6 × 4C0 = 28 ways M1 Summing the number of ways for 2 or 3 correct scenarios (can be unsimplified), no incorrect scenarios Total 672 ways A1 Correct answer 3 Question Answer Marks Guidance 4(ii) Total number of selections = 12C6 = 924 (A) M1 12Cx – (subtraction seen), accept unsimplified Selections with males together = 10C4 = 210 (B) A1 Correct unsimplified expression Total = (A) – (B) = 714 A1 Correct answer Alternative method for question 4(ii) No males + Only male 1 + Only male 2 = 10C6 + 10C5 + 10C5 M1 10Cx + 2 x 10Cy , x ≠ y seen, accept unsimplified = 210 + 252 + 252 A1 Correct unsimplified expression = 714 A1 Correct answer Alternative method for question 4(ii) Pool without male 1 + Pool without male 2 – Pool without either male M1 2 x 11Cx – 10Cx = 11C6 + 11C6 – 10C6 = 462 + 462 – 210 A1 Correct unsimplified expression = 714 A1 Correct answer 3

This question in 9709/63 May/June 2019

6 (i) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that all four Es are together. [1] … … … … (ii) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that the Ss are not next to each other. [4] … … … … … … … … … … … … … … … … … … … Four letters are selected from the 12 letters of the word STEEPLECHASE. (iii) Find the number of different selections if the four letters include exactly one S. [4] … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) 9! 2! 181440 = B1 Exact value 1 6(ii) Total no of ways = 12! 9 979 200 2!4! = (A) B1 Accept unevaluated With Ss together = 11! 1663 200 4! = (B) B1 Accept unevaluated With Ss not together = (B) – (A) M1 Correct or 12! 8!, , integers 1 m n m n − > or their identified total – their identified Ss together 8 316 000 A1 Exact value Alternative method for question 6(ii) _ T _ E _ E _ P _ L _ E _ C _ H _ A _ E _ B1 10! × k in numerator k integer ⩾ 1 10! 11 10 4! 2! × × B1 4! × k in numerator k integer ⩾ 1 1 0! 4! their their × 11C2 or 11P2 M1 OE 8 316 000 A1 Exact value 4 Question Answer Marks Guidance 6(iii) S E E E : 1 M1 6Cx seen alone or times K > 1 S E E _ : 6C1 = 6 S E _ _ : 6C2 = 15 S _ _ _ : 6C3 = 20 B1 6C3 or 6C2 or 6C1 alone Add 3 or 4 correct scenarios M1 No extras Total = 42 A1 4

This question in 9709/61 Oct/Nov 2019

7 (i) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that all three Os are together and both Ts are together. [1] … … … … … … (ii) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that the Ts are not together. [4] … … … … … … … … … … … … … … … … … (iii) Find the probability that a randomly chosen arrangement of the 9 letters of the word TOADSTOOL has a T at the beginning and a T at the end. [2] … … … … … … … … (iv) Five letters are selected from the 9 letters of the word TOADSTOOL. Find the number of different selections if the five letters include at least 2 Os and at least 1 T. [4] … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(i) B1 Evaluated 1 7(ii) Total no of arrangements: 9! 2!3! 30240 = B1 Accept unevaluated No with Ts together = 8! 3! 6720 = B1 Accept unevaluated With Ts not together: 30 240 – 6720 M1 correct or 9! 8!, , integers 1 m n m n − > or their identified total – their identified Ts together 23 520 A1 CAO Alternative method for question 7(ii) 7! 8 7 3! 2 × × B1 7! × (k > 0) in numerator, cannot be implied by 7P2, etc. B1 3! × (k > 0) in denominator M1 7! 3! their their × 8C2 or 8P2 23 520 A1 CAO 4 Question Answer Marks Guidance 7(iii) Number of arrangements = 7! 3! Probability = 7! 840 3! 9! 30240 3!2! their their = M1 identified number of arrangements with T at ends identified total number of arrangements their their 7! , integers 1 9! m or m n n > 1 36 or 0.0278 A1 Final answer 2 7(iv) OOT_ _ 4C2 = 6 OOTT_ 4C1 = 4 OOOT_ 4C1 = 4 OOOTT = 1 M1 4Cx seen alone or 4Cx x k ≥1, k an integer, 0< x <4 A1 4C2 x k, k = 1 oe or 4C1 x m, m = 1 oe alone M1 Add 3 or 4 identified correct scenarios only, accept unsimplified (Total) = 15 A1 CAO, WWW Only dependent on 2nd M mark 4

This question in 9709/62 Oct/Nov 2019

2 (i) How many different arrangements are there of the 9 letters in the word CORRIDORS? [2] … … … … … … … … … (ii) How many different arrangements are there of the 9 letters in the word CORRIDORS in which the first letter is D and the last letter is R or O? [3] … … … … … … … … … … … … … …

5 marks

Mark scheme: 2(i) 9! 2!3! = 30240 B1 9! Divided by at least one of 2! or 3! B1 Exact value 2 2(ii) D _ _ _ _ _ _ _ R: 7! 2!2!= 1260 D _ _ _ _ _ _ _ O: 7! 3! = 840 B1 7! Seen alone or as numerator in a term, can be multiplied not + or – B1 One term correct, unsimplified Total = 2100 B1 Final answer 3

This question in 9709/63 Oct/Nov 2019

3 A sports team of 7 people is to be chosen from 6 attackers, 5 defenders and 4 midfielders. The team must include at least 3 attackers, at least 2 defenders and at least 1 midfielder. (i) In how many different ways can the team of 7 people be chosen? [4] … … … … … … … … … … … … … The team of 7 that is chosen travels to a match in two cars. A group of 4 travel in one car and a group of 3 travel in the other car. (ii) In how many different ways can the team of 7 be divided into a group of 4 and a group of 3? [2] … … … … … … … …

6 marks

Mark scheme: 3(i) 4A 2D 1M : 6C4 × 5C2 × 4C1 (= 600) 3A 3D 1M : 6C3 × 5C3 × 4C1 (= 800) M1 A1 2 correct products, allow unsimplified M1 Summing their totals for 3 correct scenarios only Total = 2600 A1 Correct answer SC1 6C3 × 5C2 × 4C1 × 9C1 = 7200 4 Question Answer Marks Guidance 3(ii) 7C4 × 1 B1 7C3 or 7C4 seen anywhere 35 B1 2

This question in 9709/63 Oct/Nov 2019

2 A club has 264 members, numbered from 1 to 264. Donash wants to choose a random sample of members for a survey. In order to choose the members for the sample he uses his calculator to generate random digits. His first 20 random digits are as follows. 10612 11801 21473 22759 (a) The numbers of the first two members in the sample are 106 and 121. Write down the numbers of the next two members in the sample. [2] … … … … … … … … … … (b) To obtain the numbers for members after the 4th member, Donash starts with the second random digit, 0, and obtains the numbers 061 and 211. Explain why this method will not produce a random sample. [1] … … … … … … … …

3 marks

Mark scheme: 2(a) 180, 227 B1 One correct. Ignore incorrect numbers. B1 Both correct and no extra numbers seen. (Allow other correct use of list of digits). 2 2(b) These numbers are not independent of the previous numbers OR Only a finite number of digits used B1 Already used these numbers, so therefore not random. Does not include numbers not in the list, therefore not random (not random or biased needs a reason). 1

This question in 9709/63 May/June 2023