Permutations and combinations — Paper 5 — A Level Mathematics 9709 topical questions

Question 1

The 40 members of a club include Ranuf and Saed. All 40 members will travel to a concert. 35 members will travel in a coach and the other 5…

Accepted Answer

Question Answer Marks Guidance 1 38Cr or nC34 M1 Either expression seen OE, no other terms, condone x1 38C34 A1 Correct unsimplified OE 73815 A1 If M0, SCB1 38C34 x k, k an integer 3

Question 2

Richard has 3 blue candles, 2 red candles and 6 green candles. The candles are identical apart from their colours. He arranges the 11 candl…

Accepted Answer

4(a) R ^ ^ ^ ^ ^ ^ ^ ^ ^ R M1 9! Alone on numerator, 9! 3! × k or 6! × k on denominator 3!6! = 84 A1 2 4(b) ^ (B B B) ^ ^ ^ ^ ^ M1 7! × k or 7k seen, k an integer > 0 6! 7! 8 × 7 M1 m × n ( n –1) or m × n C 2 or m × n P2 , n=7, 8 or 9, m an integer > 0 × 6! 2 M1 n = 8 used in above expression = 196 A1 Alternative for question 4(b) [Arrangements, blues together – Arrangements with blues M1 9! Seen alone or as numerator with subtraction together and reds together =] 9! 8! − 2!6! 6! = [252 – 56] M1 8! Seen alone or as numerator in a second term and no other terms M1 All terms divided by 6! x k, k an integer = 196 A1 4

Question 3

(a) Find the number of diﬀerent arrangements that can be made from the 9 letters of the word JEWELLERY in which the three Es are together a…

Accepted Answer

2(a) 6! M1 720 A1 2 2(b) Total number: ( ) 9! 30240 3!2! M1 Number with Ls together = ( ) 8! 6720 3! M1 Number with Ls not together = 9! 8! 3!2! 3! − = 30 240 – 6720 M1 23 520 A1 Alternative method for question 2(b) 7! 8 7 3! 2 × × 7! × k in numerator, k integer ≥ 1 M1 8 × 7 × m in numerator or 8C2 × m, m integer ≥ 1 M1 3! in denominator M1 23 520 A1 4

Question 4

In a music competition, there are 8 pianists, 4 guitarists and 6 violinists. 7 of these musicians will be selected to go through to the ﬁna…

Accepted Answer

4 Scenarios: 2P 3V 2G 8C2 × 4C2 × 6C3 = 28 × 6 × 20 = 3360 2P 4V 1G 8C2 × 4C1 × 6C4 = 28 × 4 × 15 = 1680 3P 3V 1G 8C3 × 4C1 × 6C3 = 56 × 4 × 20 = 4480 4P 2V 1G 8C4 × 4C1 × 6C2 = 70 × 4 × 15 = 4200 (M1 for 8Cr × 4Cr × 6Cr with ∑ݎ = 7) Two unsimplified products correct B1 Summing the number of ways for 3 or 4 correct scenarios M1 Total: 13 720 A1 4

Question 5

(a) Find the number of diﬀerent ways in which the 10 letters of the word SUMMERTIME can be arranged so that there is an E at the beginning …

Accepted Answer

6(a) 8! 3! M1 6720 A1 2 Question Answer Marks 6(b) Total number = ( ) ( ) 10! 302400 A 2!3! B1 With Es together = ( ) ( ) 9! 60480 B 3! B1 Es not together = their (A) – their (B) M1 241920 A1 Alternative method for question 6(b) _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ 8! 9 8 3! 2 × × 8! × k in numerator, k integer ≥ 1, denominator ≥ 1 B1 3! × m in denominator, m integer ≥ 1 B1 Their 8! 3! Multiplied by 9C2 (OE) only (no additional terms) M1 241920 A1 4 Question Answer Marks 6(c) Scenarios: E M M M 5C0 = 1 E M M _ 5C1 = 5 E M _ _ 5C2 = 10 M1 Summing the number of ways for 2 or 3 correct scenarios M1 Total = 16 A1 3

Question 6

(a) Find the number of diﬀerent possible arrangements of the 9 letters in the word CELESTIAL. [1] .........................................…

Accepted Answer

7(a) 9! 2!2!= 90 720 B1 1 7(b) 6! 2! M1 360 A1 2 Question Answer Marks 7(c) 2 Es together = ( ) 8! 20160 2! = M1 Es not together = 90720 – 20160 = 70560 M1 Probability = 70560 90720 M1 7 or 0.778 9 A1 Alternative method for question 7(c) _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ 7! 8 7 2! 2 × × = 70560 7! × k in numerator, k integer ⩾ 1, denominator ⩾ 1 M1 Multiplying by 8C2 OE M1 Probability = 70560 90720 M1 7 or 0.778 9 A1 4 Question Answer Marks 7(d) Scenarios are: E L _ _ _ 5C3 10 E E L _ _ 5C2 10 E _ _ _ _ 5C4 5 E E _ _ _ 5C3 10 M1 Summing the number of ways for 3 or 4 correct scenarios M1 Total = 35 A1 3

Question 7

(a) Find the number of diﬀerent ways in which the 10 letters of the word SHOPKEEPER can be arranged so that all 3 Es are together. [2] ....…

Accepted Answer

7(a) 8! 2! M1 ( ) 8! 7! 8 ! ,where , , where 2 ! a k a k k × ≡ ∈ ∈   20160 A1 2 Question Answer Marks Guidance 7(b) Total number of ways: ( ) 10! 302400 2!3! = (A) B1 Accept unsimplified With Ps together: 9! ( 3! = 60 480) (B) B1 Accept unsimplified With Ps not together: 302 400 – 60 480 M1 10! 9! m n − , m, n integers or (A) – (B) if clearly identified 241 920 A1 Alternative method for question 7(b) 8! 3! B1 k × 8! in numerator, k a positive integer, no ± B1 m × 3! in denominator, m a positive integer, no ± 9 8 2 × × M1 Their 8! 3!multiplied by 9C2 or 9P2 no additional terms 241 920 A1 Exact value, WWW 4 Question Answer Marks Guidance 7(c) Probability = Number of ways Es at beginning and end Total number of ways Probability = 8! 2! 10! 2! 3! × = 20160 302400 M1 8! ! 10! ! ! k k l      1 ⩽ k, l ∈ ℕ ⩽ 3, FT denominator from 7(b) or correct 1 15 , 0·0667 A1 Alternative method for question 7(c) Probability = 3 2 10 9 × M1 1 3,2 10 9 a a a − × = 1 15 , 0·0667 A1 Alternative method for question 7(c) Probability = 1 1 3! 10 9 × × M1 1 1 !, 3,2 10 9 m m × × = 1 15 , 0.0667 A1 2 Question Answer Marks Guidance 7(d) Scenarios: P E E E 5C0 = 1 P E E _ 5C1 = 5 P E _ _ 5C2 = 10 P _ _ _ 5C3 = 10 M1 5Cx seen alone, 1 ⩽ x ⩽ 4 M1 Summing the number of ways for 3 or 4 correct scenarios (can be unsimplified), no incorrect scenarios Total = 26 A1 3

Question 8

Mr and Mrs Ahmed with their two children, and Mr and Mrs Baker with their three children, are visiting an activity centre together. They wi…

Accepted Answer

6(a) Condone 9C6 + 3C3, 9P6 × 3P3 84 A1 Accept unevaluated. 2 6(b) Number with 3 Baker children = 6C2 or 15 B1 Correct seen anywhere, not multiplied or added Total no of selections = 9C5 or 126 Probability = number of selections with 3 Baker children total number of selections M1 Seen as denominator of fraction 15 126 , 0·119 A1 OE, e.g. 5 42 Alternative method for question 6(b) 5 3 3 2 1 6 5 9 8 7 6 5    × × × × ×       C B1 5C3 (OE) or 10 seen anywhere, multiplied by fractions only, not added M1 3 2 1 6 5 9 8 7 6 5    × × × × ×       k , 1 ⩽ k, k integer 15 126 , 0·119 A1 OE, e.g. 5 42 3 Question Answer Marks Guidance 6(c) [Total no of arrangements = 9!] [Arrangements with men together = 8! × 2] Not together: 9! – M1 9! – k or 362880 – k, k an integer<362 880 8! × 2 B1 8! × 2(!) or 80 640 seen anywhere 282 240 A1 Exact value Alternative method for question 6(c) 7! × 8 × 7 B1 7! × k, k positive integer > 1 M1 m × 8 × 7, m × 8P2, m × 8C2 m positive integer > 1 282 240 A1 Exact value 3 6(d) 7! × 2 × 7 M1 7! × k, k positive integer > 1 If 7! not seen, condone 7 × 6 × 5 × 4 × 3 × 2 × (1) × k or 7 × 6! × k only M1 m × 2 × 7, m positive integer > 1 70 560 A1 3

Question 9

A committee of 6 people is to be chosen from 9 women and 5 men. (a) Find the number of ways in which the 6 people can be chosen if there mu…

Accepted Answer

3(a) Scenarios: 6W 0M 9C6 = 84 5W 1M 9C5 × 5C1 = 126 × 5 = 630 4W 2M 9C4 × 5C2 = 126 × 10 = 1260 M1 Correct number of ways for either 5 or 4 women, accept unsimplified M1 Summing the number of ways for 2 or 3 correct scenarios (can be unsimplified), no incorrect scenarios. Total = 1974 A1 3 3(b) Total number of ways = 14C6 (3003) Number with sister and brother = 12C4 (495) Number required = 14C6 – M1 14C6 – a value 12C4 = 3003 – 495 M1 12Cx or nC4 seen on its own or subtracted from their total, x ⩽ 6, n ⩽ 13 2508 A1 Alternative method for question 3(b) Number of ways with neither = 12C6 = 924 M1 12C6 + a value Number of ways with either brother or sister (not both) = 12C5 × 2 (= 792 × 2) = 1584 M1 12Cx × 2 or nC5 × 2 seen on its own or added to their number of ways with neither, x ⩽ 5, n ⩽ 12 Number required = 924 + 1584 = 2508 A1 3

Question 10

The 8 letters in the word RESERVED are arranged in a random order. (a) Find the probability that the arrangement has V as the ﬁrst letter a…

Accepted Answer

5(a) Total number of ways = 8! 3!2! (= 3360) B1 Correct unsimplified expression for total number of ways Number of ways with V and E in correct positions = 6! 2! 2! × (= 180) B1 6! 2! 2! × alone or as numerator in an attempt to find the number of ways with V and E in correct positions. No , × ± Probability = 180 3360 3 56   =     or 0.0536 B1 FT Final answer from their 6! 2! 2! × divided by their total number of ways Alternative method for question 5(a) 1 3 8 7 × M1 8 7 × a b seen, no other terms (correct denominators) M1 1 3 × c d seen, no other terms (correct numerators) 3 56 or 0.0536 A1 3 Question Answer Marks Guidance 5(b) Rs together and Es together: 5! (120) B1 Alone or as numerator of probability to represent the number of ways with Rs and Es together, no ×, +, – Es together: ( ) 6! 360 2! = B1 Alone or as denominator of probability to represent the number of ways with Es together, no ×, + or – Probability = 5! 6! 2! M1 5! 6! 2! their their seen 1 3 A1 OE Alternative method for question 5(b) P(Rs together and Es together): 5! 1 total number of ways 28 their   =     B1 P(Es together): 6! 3 2! 28 total number of ways their   =     B1 Alone or as numerator of probability to represent the P(Rs and Es together), no ×, +, – Probability = 1 28 3 28 M1 Alone or as denominator of probability to represent the P(Es together), no ×, + or – 1 3 A1 OE, 1 28 3 28 their their seen 4

Question 11

(a) Find the total number of diﬀerent arrangements of the 11 letters in the word CATERPILLAR. [2] .........................................…

Accepted Answer

6(a) 11! 2!2!2! 2! × m! × n! on denominator, m = 1, 2, n = 1, 2. no additional terms, no additional operations. 4989600 A1 Exact answer only. 2 Question Answer Marks Guidance 6(b) Method 1 R ^ ^ ^ ^ ^ ^ ^ R Arrange the 7 letters CTEPILL = 7! 2! Number of ways of placing As in non-adjacent places = 8 2 C 8 2 7! 2!× C B1 7! 2!× k seen, k an integer > 1. M1 ( )1 × − m n n or 2 × n m C or 2 × n m P , n = 7, 8 or 9, m an integer > 1. M1 8 2 7! !× C p or 8 2 7! !× P p , p integer ⩾ 1, condone 2520×28. = 70560 A1 Exact answer only. SC B1 70560 from M0, M1 only. Method 2 [Arrangements Rs at ends – Arrangements Rs at ends and As together] Total arrangements with R at beg. and end = 9! 2!2! Arrangements with R at ends and As together = 8! 2! With As not together = 9! 8! 2!2! 2! − M1 9! 2! ! m – k, 90720 > k integer > 1, m = 1, 2. B1 s – 8! 2! , s an integer >1 M1 9! 8! − p q , p, q integers ⩾ 1, condone 90720 – 20160. [90720 – 20160] = 70560 A1 Exact answer only. SC B1 70560 from M0, M1 only. 4 Question Answer Marks Guidance 6(c) Method 1 R R A L _ _ 5C2 = 10 R R A L L _ 5C1 = 5 R R A A L _ 5C1 = 5 R R A A L L = 1 M1 5Cx seen alone or 5Cx × k, 2⩾ k ⩾ 1, k an integer, 0 < x < 5 linked to an appropriate scenario. A1 5C2 × k, k = 1 oe or 5C1 × m, m = 1,2 oe alone. SC if 5Cx not seen. B2 for 5 or 10 linked to the appropriate scenario WWW. M1 Add outcomes from 3 or 4 identified correct scenarios only, accept unsimplified. 2Cw × 2Cx × 2Cy × 5Cz, w+x+y+z=6 identifies w Rs, × As and y Ls. [Total =] 21 A1 WWW, only dependent on 2nd M mark. Note: 5C2 + 5C1 + 5C1 + 1 = 21 is sufficient for 4/4. SC not all (or no) scenarios identified. B1 10 + 5 + 5 + 1 DB1 = 21 Method 2 – Fixing RRAL first. N.B. No other scenarios can be present anywhere in solution. R R A L ^ ^ = 7C2 M1 7Cx seen alone or 7Cx × k, 2⩾ k⩾ 1, k an integer, 0

Question 12

A bag contains 12 marbles, each of a diﬀerent size. 8 of the marbles are red and 4 of the marbles are blue. How many diﬀerent selections of…

Accepted Answer

1 BBBBR 8C1 × 4C4 = 8 RRRRR 8C5 = 56 M1 0 ⩽ y ⩽ 4 condone 8C1 × 1 A1 Two correct outcomes evaluated M1 Add 2 or 3 identified correct scenarios only (no additional terms, not probabilities) [Total =] 344 A1 WWW, only dependent on 2nd M mark 4 SC not all (or no) scenarios identified B1 280 + 8 + 56 DB1 344

Question 13

(a) How many diﬀerent arrangements are there of the 8 letters in the word RELEASED? [1] ...................................................…

Accepted Answer

3(a) 8! 3!       = 6720 1 3(b) _ _ _ L E D _ _ : With LED together: 6! 2! M1 6! k or 5! 6 x k k ⩾ 1 and no other terms M1 2! m , m an integer, m ⩾ 5 360 A1 CAO 3 3(c) Method using _ _ _ A _ D _ _ : Arrange the 6 letters RELESE = 6! 3! [= 120] *M1 6! 3!×k seen, k an integer > 0 Multiply by number of ways of placing AD in non-adjacent places = their 120 × 7P2 [= 5040] *M1 ( ) 1 × − m n n or 2 × n m C or 2 × n m P , n = 6, 7 or 8, m an integer > 0 [Probability =] 5040 6720 their their DM1 Denominator = their (a) or correct, dependent on at least one M mark already gained. 5040 3 or or 0.75 6720 4 A1 Alternative method for Question 3(c) Method using ‘Total arrangements – Arrangements with A and D together’: Their 6720 – 7! 2 3! × [= 5040] *M1 Their 6720 – k, k a positive integer *M1 ( ) 7! , 1,2 3! × − = k m k Question Answer Marks Guidance [Probability =] 5040 6720 their their DM1 With denominator = their (a) or correct, dependent on at least one M mark already gained. 5040 3 or or 0.75 6720 4 A1 Alternative method for Question 3(c) Method using ‘1 – Probability of arrangements with A and D together’: 7! 2 3! × [= 1680] *M1 7! , 1,2 3! × = k k [Probability =] 1 680 6720 their their *M1 With denominator = their (a) or correct 1 – 1 680 6720 their their DM1 1 – m, 0 < m < 1 , dependent on at least one M mark already gained 5040 3 or or 0.75 6720 4 A1 4

Question 14

(a) Find the total number of diﬀerent arrangements of the 8 letters in the word TOMORROW. [2] .............................................…

Accepted Answer

6(a) 8! 2!3! M1 8! ! ! × k m k = 1 or 2, m = 1 or 3, not k = m = 1 no additional terms 3360 A1 2 Question Answer Marks Guidance 6(b) Method 1 Arrangements Rs at ends – Arrangements Rs at ends and Os together [Os not together = ] 6! 3! – 4! M1 6! ! k – m, 1 ⩽ k ⩽ 3, m an integer, condone 6! 2 !   × −     m k . M1 w – 4! or w – 24, w an integer Condone w – 2 × 4! 96 A1 Method 2 identified scenarios R _ _ _ R, Arrangement No Os together + 2Os and a single O 4C3 × 3! + 4C2 × 2 × 3! M1 4C3 × 3! + r or 4× 3! + r or 4P3 × 3! + r, r an integer. Condone 2 × 4C3 × 3! + r. 2 × 4× 3! + r or 2 × 4P3 × 3! + r. M1 q + 4C2 × 3! × k or q + 4P2 × 3! × k, k = 1,2, q an integer [24 + 72 =] 96 A1 3 6(c) Method 1 Identified scenarios OORR 3 2 3 2 2 0 C C C 3 1 3   × × = × =   ORR_ 3 2 3 1 2 1 C C C 3 1 3 9 × × = × × = OOR_ 3 2 3 2 1 1 C C C 3 2 3 18 × × = × × = OR_ _ 3 2 3 1 1 2 C C C 3 2 3 18 × × = × × = OOOR 3 2 3 3 1 0 C C C 1 2 2   × × = × =   B1 Outcomes for 2 identifiable scenarios correct, accept unsimplified. M1 Add 4 or 5 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. Total 50 A1 All correct and added Probability = 8 4 50 C M1 8 4 '50' their C , accept numerator unevaluated Question Answer Marks Guidance 6(c) cont’d 50 or 0.714 70 A1 Method 2 Identified outcomes ORTM 3 2 1 1 C C × = 6 ORTW 3 2 1 1 C C × = 6 ORMW 3 2 1 1 C C × = 6 ORRM 3 2 1 2 C C × = 3 ORRW 3 2 1 2 C C × = 3 ORRT 3 2 1 2 C C × = 3 OROR 3 2 2 2 C C × = 3 OROT 3 2 2 1 C C × = 6 OROM 3 2 2 1 C C × = 6 OROW 3 2 2 1 C C × = 6 OROO 3 2 3 1 C C × = 2 B1 Outcomes for 5 identifiable scenarios correct, accept unsimplified. M1 Add 9, 10 or 11 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. Total 50 A1 All correct and added Probability = 8 4 50 C M1 8 4 '50' their C , accept numerator unevaluated. 50 or 0.714 70 A1 5

Question 15

(a) How many diﬀerent arrangements are there of the 11 letters in the word REQUIREMENT? [2] ...............................................…

Accepted Answer

6(a) 11! 2!3! M1 11! alone on numerator – must be a fraction. k! × m! on denominator, k = 1, 2, m = 1, 3, 1 can be implied but cannot both = 1. No additional terms 3326400 A1 Exact value only 2 6(b) 8! = 40320 B1 Evaluate, exact value only 1 6(c) 9! 7 3!× M1 9! 3!×k seen, k an integer > 0, no +, – or ÷ M1 7 × an integer seen in final answer, no +, – or ÷ 423360 A1 Exact value only Alternative method for Question 6(c) 9C3 ×7! (× 3! 3! ) M1 9C3×k seen, k an integer > 0, no + or – M1 7! × k seen, , k an integer > 0, no + or – 423360 A1 Exact value only but there must be evidence of 3! 3! × Question Answer Marks Guidance 6(c) cont’d Alternative method for Question 6(c) 8! 3 7 2! × × M1 8! 3 2! × ×k seen, k an integer > 0, no + or – M1 7 × an integer seen in final answer, no +, – or ÷ 423360 A1 Exact value only Alternative method for Question 6(c) 2 9 8 7 1 7 total no. of arrangements 11 10 9 8 7 × × × × × × M1 Product of correct five fractions × k seen, k an integer > 0, no + or – M1 7×’total no of arrangements’ ×k seen, k an integer > 0, no + or – 423360 A1 Exact value only Alternative method for Question 6(c) No E between the Rs – 6 3 3! 7! 100800 3! C × × = 1E between the Rs – 6 2 3! 7! 226800 2! C × × = 2Es between the Rs – 6 1 3! 7! 90720 C × × = 3Es between the Rs – 7! = 5040 M1 Finding the correct number of ways for no, 1 or 2 Es between the Rs, accept unsimplified. M1 Adding the number of ways for 3 or 4 correct scenarios ( ) Total 7! 20 45 18 1 7! 84 423360  = × + + + = × =   A1 CAO 3 Question Answer Marks Guidance 6(d) E E R _ _ 6C2 = 15 E E R R _ 6C1 = 6 E E E R _ 6C1 = 6 E E E R R 6C0 = 1 M1 Identifying four correct scenarios only. B1 Correct number of selections unsimplified for 2 or more scenario. M1 Adding the number of selections for 3 or 4 identified correct scenarios only, accept unsimplified. 3Cx ×2Cy × 6Cz, x+y+z=5 correctly identifies x Es and y Rs [Total =] 28 A1 WWW, only dependent upon 2nd M mark. Alternative method for Question 6(d) – Fixing EER first. No other scenarios can be present anywhere in solution. E E R ^ ^ = 8C2 M1 8Cx seen alone or 8Cx × k, , k = 1 or 2, 0

Question 16

Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken. For the ﬁrst ph…

Accepted Answer

5(a) [8! =] 40 320 B1 Evaluated, exact value only. 1 5(b) Method 1 [^ ^ ^ R ^ ^ S ^ ^] 7! × 8C2 × 2 M1 7! × k seen, k an integer > 1. M1 ( ) 1 × − m n n or 2 × n m C or 2 × n m P , n = 7, 8 or 9, m an integer > 1. 282 240 A1 Exact value only. SC B1 for final answer 282 240 WWW. Method 2 [Total number of arrangements – Arrangements with R & S together] 9! – 8! × 2 M1 9! – k, k an integer < 362 880 . M1 m – 8! × n, m an integer > 40 320, n = 1,2. 282 240 A1 Exact value only. SC B1 for final answer 282 240 WWW. 3 5(c) 9C5 [× 4C4] M1 9Cx [× 9–xC9–x,] x = 4, 5. Condone × 1 for 9–xC9–x. Condone use of P. 126 A1 WWW 2 Question Answer Marks Guidance 5(d) [Number of ways with Raman and Sanjay together on back row =] 7C3 [Number of ways with Raman and Sanjay together on front row =] 7C2 M1 7Cx seen, x = 3 or 2. [Total =] 35 + 21 M1 Summing two correct scenarios. 56 A1 Evaluated – may be seen used in probability. If M0 scored, SC B1 for 56 WWW. Probability = ( ) 56 56 4 , 126 9 = their their c , 0.444 B1 FT FT their 56 from adding 2 or more scenarios in numerator and their (c) or correct as denominator. 4

Question 17

A group of 6 people is to be chosen from 4 men and 11 women. (a) In how many diﬀerent ways can a group of 6 be chosen if it must contain ex…

Accepted Answer

2(a) M1 1848 A1 CAO as exact. 2 2(b) Method 1 [Identifying scenarios] [Neither selected =] 13C6 [= 1716] [Only Jane selected =] 13C5 [= 1287] [Only Kate selected =] 13C5 [= 1287] M1 Either 13C6 seen alone or 13C5 seen alone or × 2 (condone 13Pn, n = 5,6). [Total =] 1716 + 1287 + 1287 M1 Three correct scenarios only added, accept unsimplified (values may be incorrect). 4290 A1 Method 2 [Total number of selections – selections with Jane and Kate both picked] 15C6 - 13C4 [= 5005 – 715] M1 15C6 – k, k a positive integer < 5005, condone 15P6. M1 m – 13C4, m integer > 715, condone n – 13P4, n > 17 160. 4290 A1 3 SC Where the condition of 2(a) is also applied in 2(b), the final answer is 1512 SC M1 M1 A0 max. The method marks can be earned for the equivalent stages in each method. Method 1 4C1 × 9C5 + 4C1 × 9C4 × 2 Method 2 4C1 × 11C5 – 4C1 × 9C3

Question 18

(a) In how many diﬀerent ways can the 9 letters of the word TELESCOPE be arranged? [2] ....................................................…

Accepted Answer

4(a) 9! 3! M1 9!,! e e = 2, 3 60 480 A1 2 Question Answer Marks Guidance 4(b) 7! 2 6 3!× × M1 7! 3! k × seen, k an integer > 0. M1 ! 2 ! m q n × × 7 ⩽ m ⩽ 9, 1 ⩽ n ⩽ 3, 1 ⩽ q ⩽ 8 all integers. M1 ! 6 ! m p n × × 7 ⩽ m ⩽ 9, 1 ⩽ n ⩽ 3, 1 ⩽ p ⩽ 2 all integers. (Accept 3P2 for 6) If M0 M0 M0 awarded, SC M1 for t × 12, t an integer ⩾ 20, 5! 3! . 10 080 A1 Exact value. Alternative method for question 4(b) 7 2P 6! 2 3! × × M1 6! 3! k × seen, k an integer > 0. M1 7 2 ! P ! m q n × × m = 6,9, 1 ⩽ n ⩽ 3, 1 ⩽ q ⩽ 2 all integers. M1 ! ! m n × 7Pr × 2 m = 6, 9, 1 ⩽ n ⩽ 3, 1 ⩽ r ⩽ 5 all integers. If M0 M0 M0 awarded, SC M1 for t × 84, t an integer ⩾ 20, 5! 3!. 10 080 A1 Exact value.

Question 19

The 26 members of the local sports club include Mr and Mrs Khan and their son Abad. The club is holding a party to celebrate Abad’s birthda…

Accepted Answer

1 M1 100947 A1 CAO 2

Question 20

A security code consists of 2 letters followed by a 4-digit number. The letters are chosen from {A, B, C, D, E} and the digits are chosen f…

Accepted Answer

5(a) M1 16 800 A1 2 Question Answer Marks Guidance 5(b) Method 1 [Identify scenarios] With A and no 5: 8 × 6P4 or (1 × 4 × 6 ×5 × 4 × 3) ×2 or 4C1 × 2! × 6P4 = 2880 With 5 and no A: 4P2 × 4 × 6P3 or (4 × 3 × 1 × 6 × 5 × 4) × 4 or 4P2 × 6C3 × 4! = 5760 With A and 5: 8× 4 × 6P3 or (4 × 1 × 1 × 6 × 5 × 4) × 8 or 4C1 × 2! × 6C3 × 4! = 3840 M1 One number of ways correct, accept unsimplified. M1 Add 2 or 3 identified correct scenarios only, accept unsimplified. [Total =] 12 480 A1 CAO Method 2 [total number of codes – number of codes with no A or 5] No A or 5 : (4 × 3 ) ×( 6 × 5 × 4 × 3) = 4320 M1 4P2 × 6P4 or 4C2 × 6C4 seen, accept unsimplified. Required number = their (a) – their 4320 M1 Their 5(a) (or correct) – their (No A or 5) value. 12 480 A1 Method 3 [subtracting double counting] With A 4P1 × 7P4 × 2 or 4C1 × 2 × 7C4 × 4! = 6720 With 5 5P2 × 6P3 × 4 or 5C2 × 2 × 6C3 × 4! = 9600 With A and 5 = 4P1 × 6P3 × 8or 4C1 × 2! × 6C3 × 4! × 8 = 3840 M1 One outcome correct, accept unsimplified. Required number = 6720 + 9600 – 3840 M1 Adding ‘with a’ to ‘with 5’ and subtracting ‘A and 5’. 12 480 A1 CAO 3 Question Answer Marks Guidance 5(c) Method 1 – number of successful codes divided by total (1 ×) 3 × 5P2 M1 3 × 5Pn, n = 2, 3. Condone 3 × 5C2, no + or –. Probability = 3 5 2 1 6 800 their P their × M1 Probability = 60 1 6 800 their their . 1 280 , 0.00357 A1 Method 2 – product of probabilities of each part of code 1 1 1 3 5 4 5 4 7 6 5 4   × × × × ×     or 1 1 3 5 2 5 4 7 4 P P × × × M1 1 1 5 4 k × × where 0 1 k < < for considering letters. M1 t 1 3 7 6 × × or 3 5 2 7 4 P t P × × where 0 1 t < < . 1 280 A1 CAO 3

Question 21

A group of 12 people consists of 3 boys, 4 girls and 5 adults. (a) In how many ways can a team of 5 people be chosen from the group if exac…

Accepted Answer

5(a) M1 Condone 5P1 for M1 only 175 A1 2 Question Answer Marks Guidance 5(b) 2B 1G 2A 3C2 × 4C1 × 5C2 = 120 2B 2G 1A 3C2 × 4C2 × 5C1 = 90 2B 3G 3C2 × 4C3 = 12 3B 1G 1A 3C3 × 4C1 × 5C1 = 20 3B 2G 3C3 × 4C2 = 6 M1 3Cx × 4Cy × 5Cz , x + y + z = 5, x,y,z integers ⩾1 Condone use of permutations for this mark B1 2 appropriate identified outcomes correct, allow unsimplified M1 Summing their values for 4 or 5 correct identified scenarios only (no repeats or additional scenarios), condone identification by unsimplified expressions [Total =] 248 A1 Note: Only dependent upon M marks 4 5(c) 8! × 3! × 5P2 M1 8! × m, m an integer ⩾ 1 Accept 8 × 7! for 8! M1 3! × n, n an integer > 1 M1 p × 5P2, p × 5C2 × 2, p × 20, p an integer > 1 If extra terms present, maximum 2/3 M marks available 4 838 400 A1 Exact value required 4

Question 22

(a) Find the number of diﬀerent arrangements of the 8 letters in the word DECEIVED in which all three Es are together and the two Ds are to…

Accepted Answer

1(a) 5! M1 k! where k = 5, 6 or 7 Condone  1 OE 120 A1 2 1(b) [Total no of ways =] 8! 2!3! [= 3360] M1 8! , 1,2 1,3 ! ! a b a b a b    [With 3Es together =] 6! 2! [= 360] M1 6!, 1,2 ! c c  seen in an addition/subtraction [With 3Es not together] = 3360 – 360 M1 8! 6! where , 1,2 & 1,3 ! ! ! d f e d e f    3000 A1 4

Question 23

There are 6 men and 8 women in a Book Club. The committee of the club consists of ﬁve of its members. Mr Lan and Mrs Lan are members of the…

Accepted Answer

2(a) M1 990 A1 Alternative method for question 2(a) [total – both on – neither on] 14C5 – ( 12C3 + 12C5) = [2002 – 220 – 792] M1 kC5 – (aC3 + aC5) a = 12, 13 and k = 13, 14 990 A1 2 2(b) [Mrs Lan plus] 2W 2M 7C2  6C2 = 315 3W 1M 7C3  6C1 = 210 4W 7C4 = 35 M1 7Cr  6C4-r for r = 2, 3 or 4 B1 Outcome for one identifiable scenario correct, accept unevaluated M1 Add outcomes for 3 identifiable correct scenarios Note: if scenarios not labelled, they may be identified by seeing7Cr  6Cs r + s = 4 to imply r women and s men for both B & M marks only [Total =] 560 A1 4

Question 24

(a) Find the number of diﬀerent arrangements of the 9 letters in the word CROCODILE. [1] ..................................................…

Accepted Answer

6(a) 9! 2!2!        90 720 1 6(b) Method 1 Arrangements Cs at ends – Arrangements Cs at ends and Os together [Os not together =] 7! 6! 2!  [= 2520 – 720] M1 ! 2! w – y, w = 6, 7 y an integer. Condone ! 2 2! w y         . M1 a – 6! or a – 720, a an integer resulting in a positive answer. 1800 A1 Method 2 identified scenarios R ^ ^ ^ R [Os not together =] 6 5 5! 2!   = M1 5! × b, b integer >1. M1 6 6 2 2 6 5 or C or or1 5 2! 2! P c        , c integer > 1. 1800 A1 3 Question Answer Marks Guidance 6(c) CCO _ 5C1 = 5 CC _ _ 5C2 = 10 OOC _ 5C1 = 5 OO _ _ 5C2 = 10 C _ _ _ 5C3 = 10 O _ _ _ 5C3 = 10 B1 Correct outcome/value for 1 identified scenario. Accept unsimplified. WWW M1 Add 5 or 6 values of appropriate scenarios only, no additional incorrect scenarios, no repeated scenarios. Accept unsimplified. Condone use of permutations. [Total =] 50 A1 3 6(d) Both Os in group with a C 5C2 = 10 Both Os in group without a C 5C2 × 3C2 = 30 One O in a C group, one not 5C1 × 4C2 = 30 One O with each C (5C1 × 4C1) ÷2! = 10 B1 A correct scenario calculated accurately. Accept unsimplified. M1 Add 3 or 4 correct scenario values, no incorrect scenarios, accept repeated scenarios. Accept unsimplified. [Total =] 80 A1 Alternative method for question 6(d) CCO O^^ ^^^ = 5C2 = 10 CC^ O^^ O^^ = 5C1 ×4C2 = 30 CC^ OO^ ^^^ = 5C1 × 4C1 = 20 B1 A correct scenario calculated accurately. Accept unsimplified. Total ways of making three groups 9 6 6 3 C C 2 2 3   = 140 140 – (their 10+ their 30+ their 20) M1 Total subtract 2 or 3 correct scenario values, no incorrect scenarios. Accept unsimplified. 80 A1 3

Question 25

A group of 15 friends visit an adventure park. The group consists of four families. • Mr and Mrs Kenny and their four children • Mr and Mrs…

Accepted Answer

7(a) M1 no + or − .__ M1 12Cs × 12–sCt [× 12-s–tCu] s = 3, 4, 5; t = 3, 4, 5 ≠ s; u = 3, 4, 5 ≠ s ,t Alternative method for question 7(a) 12! 5! 3! 4!   M1 12! by a product of three factorials. M1 ! 5! 3! 4!   n [792 × 35 =] 27 720 A1 CAO 3 Question Answer Marks Guidance 7(b) 4! (Lizo) × 6! (Kenny) × 2! (Martin) × 2! (Nantes) M1 Product involving at least 3 of 4!, 6!, 2!, 2! × 3! (orders of K, M and N) M1 w × 3! , w integer > 1. 414 720 A1 WWW CAO 3 7(c) 7C4 (adults) × 4C1 × 3C1 M1 7C4 × b, b integer > 1 no + or – . 420 A1 2 7(d) K not L 5C3 × 8C3 = 560 L not K 5C3 × 8C3 = 560 L and K 5C2 × 8C3= 560 M1 8C3(or 8P3) × c for one of the products or 5C3 (or 5P3)× c, positive integer >1 for first 2 products only. M1 Add 2 or 3 correct scenarios only values, no additional incorrect scenarios, no repeated scenarios. Accept unsimplified. [Total or Difference=] 1680 A1 Alternative method for question 7(d) Total no of ways – neither L nor K Total = 7C4 × 8C3 = 1960 Neither K nor L = 5C4  8C3 =280 M1 8C3 × c, c a positive integer >1. M1 Subtracting the number of ways with neither from their total number of ways. [Total or Difference=] 1680 A1 Question Answer Marks Guidance 7(d) Alternative method for question 7(d) Subtracting K and L from sum of K and L K 6C3 × 8C3 = 1120 L 6C3 × 8C3 = 1120 L and K 5C2 × 8C3= 560 1120 + 1120 – 560 = 1680 M1 8C3 × c, c a positive integer >1. M1 Subtracting number of ways with both from sum of number of ways with K and number of ways with L. [Total or Difference=] 1680 A1 3

Question 26

A Social Club has 15 members, of whom 8 are men and 7 are women. The committee of the club consists of 5 of its members. (a) Find the numbe…

Accepted Answer

6(a) 5M0W 8C5 [× 7 C0] = 56 M1 8Cx × 7 C5-x for x = 1, 2, 3, 4,or 5 4M1W 8C4 × 7 C1 = 490 3M2W 8C3 × 7 C2 = 1176 B1 Outcome for 4M1W or 3M2W correct and identified, accept unsimplified. M1 Add 3 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified. Addition may be implied by final answer. [Total =] 1722 A1 Value stated WWW. Alternative method for Question 6(a) 2M3W 8C2 × 7 C3 = 980 M1 8Cx × 7 C5-x for x = 1, 2, 3, 4,or 5 1M4W 8C1 × 7 C4 = 280 0M5W 8C0 × 7 C5 = 21 B1 Outcome for 2M3W or 1M4W correct and identified, accept unsimplified. [Total = 15C5 – (980 + 280 + 21)] M1 Subtract 3 values of appropriate scenarios from their identified 3003 – (980 + 280 + 21) total or correct, no incorrect scenarios, no repeated scenarios, accept unsimplified. [Total =] 1722 A1 Value stated WWW. 4 6(b) 15C3 × 12 C5 [× 7C7] [= 455 × 792 ] M1 15Cr × q, r = 3, 5, 7; q a positive integer >1 M1 15Cs × 15-sCt [× 15-s-tCu] s = 3,5,7; t = 3,5,7 ≠ s; u = 3,5,7 ≠ s,t 360360 A1 Final answer. If A0 awarded SC B1 for final answer 360360. 3 6(c) Method 1: Total number of arrangements with AB together – Arrangements with AB and FG together 6! × 2 – 5! ×2 ×2 M1 a! × 2! × b, a = 5, 6; b = 1,2 seen. [ = 1440 – 480 ] M1 Either 6! × 2 – c ¸1 < c < 1440 or d – 5! ×2 ×2, 1440 < d 960 A1 Method 2: arrangements with AB together with F and G not together. 2 × 4! × 5 × 4 M1 2 × 4! × e, e positive integer >1 M1 f × 5 × 4, f positive integer >1 condone f × 20, f × 5C2, f positive integer >1 960 A1 3

Question 27

(a) Find the number of diﬀerent arrangements of the 9 letters in the word ALLIGATOR in which the two As are together and the two Ls are tog…

Accepted Answer

7(a) 7! M1 7! b,c = 1,2 b ! c ! 2! 2! 7!   oe, no further terms present. 2! 2! 5040 A1 2 7(b) Method 1 for first 3 marks: Arrangements of 6 letters including Ls between As 5!  5  2 M1 5!  d, d integer > 1 M1 e!  f  g, e = 5, 6, 7; f = 1, 5; g = 1, 2; f ≠ g, 1 can be implicit. 1200 A1 Method 2 for first 3 marks: Number of arrangements of LL^^^^^ – number of arrangements with the Ls split by an A 6!  2 – 5!  2 M1 6!  2 – h h an integer 1 < h < 1440 M1 k – 5!  2 k an integer k > 240 1200 A1 Method 3 for first 3 marks: Alternative approaches to Method 1 ^A ^ ^ ^ ^ ^ A 5P1  1P1  5P5  1P1 = 600 M1 LL treated as a single unit. M1 1200 A1 7(b) Final 2 marks of Question 7(b)  9!  B1 Accept unsimplified. [Total number of arrangements =] = 90720   May be seen as denominator of probability.  2!2!  1200 5 B1 FT their 1200 Probability = , , 0.0132 unsimplified B1 FT if their 1200 and their 90 720 90720 378 their 90720 supported by work in this part. 5 7(c) Method 1: Scenarios identified Both As and Ls removed A _ _ _ _ 5C4 = 5 B1 1 correct, identified outcome/value AA _ _ _ 5C3 = 10 for A, AL or AAL scenario, accept unsimplified AL _ _ _ 5C3 = 10 5C5–x cannot be used in place of 5Cx AAL _ _ 5C2 = 10 M1 Add 4 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. [Total =] 35 A1 Value stated WWW. Method 2: 1 A fixed, 1 L removed No other scenarios can be present anywhere in solution A ^ ^ ^ ^ 7C4 M1 7Ch, 3 ⩽ h ⩽ 5 B1 7C4 oe, no other terms, scenario identified. [Total =] 35 A1 Value stated. Method 3: 1 A fixed, both Ls removed A ^ ^ ^ ^ = 6C4 = 15 B1 Correct outcome/value for 1 identified scenario, accept A L ^ ^ ^ = 6C3 = 20 unsimplified. WWW M1 Add 2 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. [Total =] 35 A1 Value stated. 3

Question 28

(a) Find the number of diﬀerent arrangements of the 9 letters in the word ACTIVATED. [2] ..................................................…

Accepted Answer

6(a) 9! M1 h ! , h = 7, 8, 9; j = 1, 2 2!2! 2! j ! 90720 A1 2 6(b) Arrangements with 5 letters between As + Arrangements with 6 letters between As + Arrangements with 7 letters between As 7! M1 7! With gap of 5: 2! 3 [= 7560] 2!k , k positive integer 1< k < 7 7! With gap of 6: 2 [= 5040] M1 Add their no of ways for 3 identified correct scenarios, no 2! additional incorrect scenarios, accept unsimplified. 7! With gap of 7: 2! 1 [= 2520] 7! A1 [Total no = 2!=]6 15120 3 6(c) Method 1: Summing number of ways AT _ _ _ 2×2× 5C3 40 B1 Correct no of ways for 4 correctly identified scenarios, A _ _ _ _ 2×5C4 10 accept unsimplified. AATT _ 5C1 5 AAT _ _ 2×5C2 20 M1 Add no of ways for 5 or 6 identified correct scenarios, no AA _ _ _ 5C3 10 additional incorrect scenarios, no repeated scenarios, accept _ _ _ _ _ 5C5 1 unsimplified. [Total no of ways not containing more Ts than As = ] A1 All correct and added = 40+10+5+20+10+1 [=86] 86 M1 their 86 Probability = 9 accept numerator unevaluated C 5 9C 5 ortheiridentified total 86 43 A1 , , 0.683 126 63 Method 2: Subtracting no of ways with more Ts from total T _ _ _ _ 2×5C4 10 B1 Correct no of ways for 2 correctly identified scenarios, no TTA _ _ 2×5C2 20 additional incorrect scenarios, no repeated scenarios, accept TT _ _ _ 5C3 10 unsimplified, condone use of permutations M1 Add no of ways for 2 or 3 correct scenarios and subtract from their total no of ways All correct and subtracted Total no of ways with more Ts than As =40 A1 9C5 − 40 = 86 86 M1 their 86 Probability = 9 accept numerator unevaluated C 5 9C 5 ortheiridentified total 6(c) 43 A1 , 0.683 63 5

Question 29

(a) Find the number of diﬀerent arrangements of the 9 letters in the word DELIVERED in which the three Es are together and the two Ds are n…

Accepted Answer

7(c) Scenarios B1 1 correct unsimplified outcome/value for one identified D E _ _ _ 4C3 4 scenario excluding DDEEE. D E E _ _ 4C2 6 Note: 4C1 cannot be used for 4C3 . D E E E _ 4C1 4 D D E _ _ 4C2 6 M1 Add values of 6 appropriate scenarios, no additional, incorrect D D E E _ 4C1 4 or repeated scenarios. Accept unsimplified. D D E E E [4C0] 1 [Total =] 25 A1 3

Question 30

(a) Find the number of ways in which a committee of 6 people can be chosen from 6 men and 8 women if it must include 3 men and 3 women. [2]…

Accepted Answer

2(a) M1 implied). 1120 A1 2 2(b) Method 1 0 brothers [3C0]  11C6 462 1 brother 3C1  11C5 1386 2 brothers 3C2  11C4 990 B1 3Cx  11C6 – x, with x = 1 or 2 seen. M1 Add values of 3 correct scenarios, (may be identified by the appropriate calculations) no incorrect/repeated scenarios, condone use of permutations. 2838 A1 Only dependent on the M mark. SC B1 for the correct calculation or 2838 seen WWW. Method 2 14C6 – 11C3 3003 – 165 B1 14C6 – d, where d a positive integer. M1 e – 11C3, where e is a positive integer >165. = 2838 A1 3

Question 31

(a) Find the number of diﬀerent arrangements of the 8 letters in the word COCOONED. [1] ...................................................…

Accepted Answer

3(a) 8! 2!3!        3360 1 3(b) 6! 2!2! M1 6! 2! ! f ; f = 1, 2, 3. 180 A1 2 3(c)       | P OOO CC P OOO CC P CC            5! 7! 3! M1 5! g g a positive integer, g 3360, 1. Condone numerator of 5! 3360g . M1 7! 3! h or 8! 3! h , where h is a positive integer. Condone division by 3360 in denominator. = 120 1 , , 0.143 840 7 A1 0.1428571… to at least 3SF. If M0 scored SC B1 for 1 7 WWW. 3

Question 32

In a group of 25 people there are 6 swimmers, 8 cyclists and 11 runners. Each person competes in only one of these sports. A team of 7 peop…

Accepted Answer

6(a) S + 5C + 1R 6C1  8C5  11C1 [= 6  56  11] 3696  S + 6C [+ 0R] 6C1  8C6 [ 11C0][= 6  28] = 168 B1 Correct outcome/value for 1 identified scenario, accept unsimplified, www. M1 Add values of 3 correct scenarios. No incorrect scenarios, no repeated scenarios. Condone 6Ce  8Cf  11Cg, with e + f + g = 7 to identify S, C, R. [Total =] 26964 A1 cao 4 6(b) 2!  3!  4!  6 M1 2!  3!  4!  k, k an integer > 0. 1 can be implied. =1728 A1 If A0 scored SC B1 for 1728 www. 2 Question Answer Marks Guidance 6(c) Method 1 6!  7  6  5 M1 6!  k, k an integer > 0. 1 can be implied. M1 ! ! ! m a b   7  n  r; 6 ⩽ m ⩽ 9; a = 1, 2; b = 1, 4; 1 ⩽ n, r ⩽ 6, n ≠ r. M1 ! ! ! m a b   7  6  5; 6 ⩽ m ⩽ 9; a =1, 2; b = 1, 4. 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. Method 2 6!  7P3 M1 6!  k, k an integer > 0. 1 can be implied. M1 ! ! ! m a b  × 7Pq, or ! ! ! m a b  × 7C !  q q ; 6 ≤ m ≤ 9; a =1, 2; b = 1, 4; 1 ≤ q ≤ 6. M1 ! ! ! m a b  × 7P3, or ! ! ! m a b  × 7 3 C 3!  ; 6 ≤ m ≤ 9; a =1, 2; b = 1, 4. 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. Question Answer Marks Guidance 6(c) Method 3 6!  35  3! M1 6!  k, k an integer > 0. 1 can be implied. M1 ! ! ! m a b   35  q!; 6 ⩽ m ⩽ 9; a =1, 2; b = 1, 4; 1 ⩽ q ⩽ 3. M1 ! ! ! m a b   35  6; 6 ⩽ m ⩽ 9; a =1, 2; b = 1, 4. 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. Method 4 9! – 7!3! – 3P2  6!  7  6 Or 9! – 7!3! – 3!  7!  6 [= 362 880 – 30 240 – 181 440] M1 9! – 7!r! – q, r an integer > 1, q an integer ⩽ 0. 0 and 1 may be implied. M1 ! ! ! ! s a b c   – 7!3! – q; s = 8, 9; a =1, 2; b = 1, 3; c = 1, 4; q an integer ⩾ 0. 0 and 1 may be implied. M1 ! ! ! ! s a b c   – 7!3! – 3P2  6!  6  7, 6 ⩽ s ⩽ 9, or ! ! ! ! s a b c   – 7!3! – 3!  7!  6, 6 ⩽ s ⩽ 9. a =1, 2 b = 1, 3 c = 1, 4. 1 may be implied. 151 200 A1 Condone 151 000. If A0 scored SC B1 for 151 200 www. 4

Question 33

(a) Find the number of diﬀerent arrangements of the 10 letters in the word CASABLANCA in which the two Cs are not together. [3] ...........…

Accepted Answer

7(b) AC^^^C^^^A 6! 2! × 4 6! 2! × s, with s being a positive integer. M1 ! ! t r × 4, r = 1, 2, 3 and t = 8, 7, 6. 1440 A1 Alternative Method for Question 7(b) 6 3 4 P 3! 2!   M1 6 3P 2! × k, with k being a positive integer. M1 4 × 3! × 6P ! m n , m = 2, 3 and n = 1, 2, 3. 1440 A1 3 Question Answer Marks Guidance 7(c) Scenarios AA _ _ _ 5C3 = 10 AAA _ _ 5C2 = 10 AAAA _ 5C1 = 5 B1 Correct number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www. M1 Add 3 values for 2, 3 and 4 As, no additional, incorrect or repeated scenarios. Accept unsimplified. 25 A1 Alternative Method 2 for Question 7(c) Scenarios: AAC _ _ 4C2 = 6 AA _ _ _ 4C3 = 4 AAAC _ 4C1 = 4 AAA _ _ 4C2 = 6 AAAAC 1 AAAA _ 4 B1 Correct total number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www (e.g., both values for AAC^^ and AA^^^ shown would be fine for 2As). M1 Add 6 values of appropriate scenarios only, no additional, incorrect or repeated scenarios. Accept unsimplified. 25 A1 3

Question 34

Table X Table Y In a restaurant, the tables are rectangular. Each table seats four people: two along each of the longer sides of the table …

Accepted Answer

6(a) 5C2 × 2 M1 5C2 × r, r = positive integer, 1 implied, no addition. M1 s × 2, s = 5C2 or 5P2 or if 5C2 or 5P2 not present, s = a single integer > 1 or t! × 2, 2 ⩽ t ⩽ 8, no other terms. 20 A1 3 6(b) Method 1 6C2 × 2 × 2× 2 × 4! M1 6C2 × 2 × 2× 2 × t, t = positive integer ⩾ 1. 6P2 × 2 × 2 × t, t = positive integer ⩾ 1. M1 u × 4!, u = positive integer > 1. 2880 A1 If A0 scored, SC B1 for 2880 nfww. Method 2 6! × 2 × 2 M1 6! × v, v = positive integer ⩾ 1. M1 w × 2 × 2, w = positive integer > 1. condone w × 4, w = positive integer > 1. 2880 A1 If A0 scored, SC B1 for 2880 nfww. 3 6(c) Method 1: Number of arrangements with Rajid and Sue together – Number of arrangements with Rajid and Sue together and at end of line 7! −2 6!  4 M1 7! × 2 – a, a = positive integer > 1. M1 b – 6! × 4, b = positive integer > 2880. M1 7! × c – 6! × d, c = 1,2 and d = 1, 4. = 7200 A1 If A0 scored, SC B1 for 7200 nfww. Method 2: Arrangements of 6 people and then place Rajid and Sue 6!2 5 M1 6! × e × f, e, f = positive integers ⩾ 1. M1 6! × 2 × f, f = positive integer ⩾ 1. If 5! Used, SC B1 5! × 2 × f, f = positive integer > 1. M1 6! × e × 5, e = positive integer ⩾ 1. 7200 A1 If A0, scored SC B1 for 7200 nfww. Method 3: Friends at ends picked first F ^ RS ^ ^ ^ F 6P2 × 5! × 2 M1 6P2 × e× f, e, f = positive integers ⩾ 1. M1 6P2 × 5! × f, f = positive integer ⩾ 1. Condone 6C2 × 5! × f, f = positive integer ⩾ 1. M1 6P2 × e × 2, e = positive integer ⩾ 1. Condone 6C2 × e × 2, e = positive integer ⩾ 1. 7200 A1 If A0 scored, SC B1 for 7200 nfww. 6(c) Method 4: RS placed in different possible positions ^ RS ^ ^ ^ ^ ^ 6P1 × 2 × 5! = 1440 M1 6Pn × a ×( 6 − n ) ! , a = positive integer, 1 ⩽ n ⩽ 5 seen once. ^ ^ RS ^ ^ ^ ^ 6P2 × 2 × 4! = 1440 ^ ^ ^ RS ^ ^ ^ 6P3 × 2 × 3! = 1440 M1 6Pn × 2 ×( 6 − n ) ! , a = positive integer, 1 ⩽ n ⩽ 5 seen at least 3 ^ ^ ^ ^ RS ^ ^ 6P4 × 2 × 2! = 1440 ^ ^ ^ ^ ^ RS ^ 6P5 × 2 × 1! = 1440 times in identified scenarios. M1 Add 5 values of appropriate scenarios only. No additional, incorrect or repeated scenarios. Accept unsimplified. 7200 A1 If A0 scored, SC B1 for 7200 nfww. 4

Question 35

(a) Find the number of diﬀerent arrangements of the 9 letters in the word ANDROMEDA in which no consonant is next to another consonant. (Th…

Accepted Answer

7(a) 5! 4! M1 5!  4!, e a positive integer, 1 can be 2!  2! e implied. No other terms on numerator. No addition etc. M1 f , f a positive integer, g = 1, 2. No 2!  g ! other terms on denominator. 720 A1 3 7(b) Method 1 Number of arrangements with A at each end – Number of arrangements with A at each end and 2 Ds together. 7! B1 7! 7 2!− 6! 2! – e, 5P − ,e e a positive integer or 0. M1 6! d – d > 720, r = 1, 2. r ,! = 1800 A1 Method 2 A ^ ^ ^ ^ ^ A and Ds inserted separately 6 P2 6  5 6 B1 5! × s, s a positive integer, 1 may be implied. 5!  or 5!  or 5!  C 2 2! 2 M1 6  5 t  , t a positive integer > 1, u =1, 2. u = 1800 A1 Method 3 Number of arrangements with As at each end and Ds placed in different scenarios. B1 Correct outcome/value for 1 identified Scenario position of first D scenario, accept unsimplified, www. A D ^ ^ ^ ^ ^ ^ A 5! × 5 600 M1 Add values of 5 correct scenarios, no A ^ D ^ ^ ^ ^ ^ A 5! × 4 480 incorrect/repeated scenarios. A ^ ^ D ^ ^ ^ ^ A 5! × 3 360 A ^ ^ ^ D ^ ^ ^ A 5! × 2 240 A ^ ^ ^ ^ D ^ ^ A 5! × 1 120 [Total =] 1800 A1 3 7(c) Method 1: M1 At least one correct unsimplified expression Scenarios for an identified scenario. A D ^ ^ 2C1  2C1  5C2 = 40 A D D ^ 2C1  [ 2C2  ] 5C1 = 10 [Total = ] 40 + 10 or 50 soi A1 www If M0 scored, SC B1 [total =]50 www. [Total number of selections =] 9C4 [= 126] B1 Accept evaluated, accept as denominator of probability expression. Do not condone 9C5 unless there is a clear explanation for selecting the letters not in the group. 50 25 B1 FT 0.396825… to at least 3SF. [Probability =] , 126 63 their attempted 40 + FT 10. Numerator must 126 be from an attempt to find the 2 appropriate scenarios and must be evaluated. 7(c) Method 2: M1 Numerator for at least one correct Scenarios unsimplified expression for an identified scenario. A D ^ ^ 2 2 5 4 4 960 20     P2 = , 2 2 5 4 12 2 2 1 5 12 9 8 7 6 3024 63 either or a b c d a b c d 4 seen, 6 ⩽ a,b,c,d ⩽ 9. A D D ^ 2 2 1 5 3P = 240 , 5     9 8 7 6 2! 3024 63 A1 2 2 5 4 12 2 2 1 5 + 12, a b c d a b c d 20 5 6 ⩽ a,b,c,d ⩽ 9. [Total Probability = ] + 1200 25 63 63 If M0 scored, SC B1 , g > 1200, or g 63 seen. B1 p q r s    present in all scenarios 9 8 7 6 t attempted, accept , t < 3024. 3024 1200 25 B1 FT 0.396825… to at least 3SF. , oe 3024 63 their attempted 960 + FT 240. Numerator 3024 must be from an attempt to find the 2 appropriate scenarios. 7(c) Method 3: selecting the A and then selecting 3 any letters and removing selections without Ds. 2C1 × (7C3 – 5C3) [= 2 × (35 – 10)] M1 a × (7C3 – 5C3), a = 1, 2. [Total = ] 50 A1 www If M0 scored, SC B1 [total =]50 www. [Total number of selections =] 9C4 [= 126] B1 Accept evaluated, accept as denominator of probability expression. Do not condone 9C5 unless there is a clear explanation for selecting the letters not in the group. 50 25 B1 FT 0.396825… to at least 3SF. [Probability =] , 126 63 their attempted 40 + FT 10. Numerator must 126 be from an attempt to find the 2 appropriate scenarios. Method 4: Listing outcomes. Either 10 correct outcomes for ADD^ listed or 40 correct outcomes for AD^^ listed M1 50 stated A1 www If M0 scored, SC B1 [total =]50 www. 126 stated or correct outcomes listed B1 50 25 B1 0.396825… to at least 3SF. [Probability =] , 126 63 their attempted 40 + FT 10. Numerator must 126 be from an attempt to find the 2 appropriate scenarios. 4

Question 36

Jai and his wife Kaz are having a party. Jai has invited ﬁve friends and each friend will bring his wife. (a) At the beginning of the party…

Accepted Answer

6(a)(i) Method 1 6!  2 6 M1 6!  a, a integer > 1. M1 b 2 6 , binteger ⩾ 1. = 46080 A1 Accurate answer required. SC B1 for 46080 if M0 M0 www. Alternative method for question 6(a)(i) 12 10 8 6 4 2 M1 c  d e f  g  h 2 ⩽ c,d,e,f,g,h (different integers) ⩽ 12 M1 Correct unsimplified. = 46080 A1 Accurate answer required. SC B1 for 46080 if M0 M0 www. 3 6(a)(ii) 5!  5!  2 2 M1 5!  5!  k,kpositive integer, 1 may be implied (no adding/subtracting). = 57600 A1 2 6(b) Method 1 probabilities of J & K being placed: 5 4  20 5  B1 Correct probability for one identified scenario. In the group of 5  = ,   12 11  132 33  M1 Denominator 12 11 for all probabilities, (1, 2 or 3 scenarios). 4 3  12 1  In the group of 4  = ,   12 11  132 11  A1 3 correct probabilities, accept unsimplified. 3 2  6 1  In the group of 3  = ,   12 11  132 22  5 4 4 3 3 2 M1 Adding probabilities for 3 correct scenarios.  +  +  12 11 12 11 12 11 19 A1 0.2878787 to at least 3SF. , 0.288 66 6(b) Method 2 number of arrangements of J & K being placed: B1 Correct value of one identified scenario seen, accept unsimplified. In the group of 5 10C3  7C4 [= 120  35 = 4200] M1 12Ca  12-aCb, a = 3, 4, 5; b = 3, 4, 5 (a ≠ b) In the group of 4 10C2  8C5 [= 45  56 = 2520] In the group of 3 10C1  9C5 [= 10  126 = 1260] [Total number of ways of arranging the 3 groups =] A1 27720 Seen alone or as denominator of probability –accept 12C5 7C4 = 792 35 = 27720 unsimplified. or 12C3  9C4 or 12C4  8C5 SC B1 if M0. 4200 + 2520 + 1260 = 7980 M1 Values of 3 correct scenarios added, accept unsimplified – or correct. 7980 19 A1 0.2878787 to at least 3SF. [Probability =] , , 0.288 27720 66 5 Note, alternative arrangement calculations possible e.g. In the group of 5 10C3  7C4 [= 120  35 = 4200] In the group of 4 10C5 5C2 [= 252 10 = 2520] In the group of 3 10C5  5C4 [= 252  5 = 1260]

Question 37

A new village social club has 10 members of whom 6 are men and 4 are women. The club committee will consist of 5 members. (a) In how many w…

Accepted Answer

6(a) 4M 1W: 6C4  4C1 = 60 M1 6Ca  4Cb with a + b = 5 seen, no extra terms. 3M 2W: 6C3  4C2 = 120 B1 Correct outcome/value for one clearly identified scenario. Accept unsimplified, www. 2M 3W: 6C2  4C3 = 60 Condone use of × 5C0. M1 Add values of three correct scenarios, no incorrect scenarios, no repeated scenarios. Condone 6Ca  4Cb with a + b = 5 to identify M, W for this mark. Total 240 A1 Not dependent on B1. If A0 scored, SC B1 for 240 www. 4 6(b) 6!  4!  2 M1 6! × 4! × k; k = 1, 2. 1 can be implied. = 34 560 A1 Cao. If M0 scored, SC B1 for 34 560 www. 2 6(c) Method 1 – Arrangements of OP in front row, 8 remaining people arranged. 8! × 3 × 2 M1 8! × g, g an integer greater than 1. M1 h! × 3 × j; h = 7, 8, 9; j = 1, 2 (1 may be implied). Condone 3C1 for 3. M1 h! × 3 × 2; h = 7, 8, 9. Condone 2C1 for 2. (Condone h! × 3! For M1M1). = 241 920 A1 If A0 Scored, SC B1 for 241 920. Method 2 – Two additional people selected for front row, front row arranged, remaining 6 people arranged in back row. 8C2 × 6! × 3! × 2 M1 8Ca × d, a = 2,6, d an integer greater than 1. M1 6! × e, e an integer greater than 1. M1 8Ca × f ! × 3! × 2 or 8Ca × f ! × 6 × 2; a = 2, 6; f = 5, 6, 7. = 241 920 A1 If A0 Scored, SCB1 for 241 920. Method 3 – Arrangements of two additional people for front row, front row arranged, remaining 6 people arranged in back row. 8P2 × 6! × 3! M1 8P2 × d, d an integer greater than 1. M1 6! × e, e an integer greater than 1. M1 8P2 × h! × 3! or 8P2 × h! × 6; h = 5, 6, 7. = 241 920 A1 If A0 Scored, SC B1 for 241 920. 6(c) Method 4 – Arrangements of 6 people for back row, front row arranged. 8P6 × 3! × 2! M1 8P6 × d, d an integer greater than 1. M1 3! × e, e an integer greater than 1. M1 8P6 × j! × 2; j = 1, 2, 3. = 241 920 A1 If A0 Scored, SC B1 for 241 920. 4

Question 38

The eight digits 1, 2, 2, 3, 4, 4, 4, 5 are arranged in a line. (a) How many different arrangements are there of these 8 digits? [1] ......…

Accepted Answer

7(a) 8! 2!3!        3360 1 7(b) Number of arrangements with 2s at the end – number of arrangements with 2s at the end and the 4s together 2 _ _ _ _ _ _ 2 – 2 _ (444) _ _ 2 6! 3! 4! M1 6! 3!  r s , r = 1, 2 and s a positive integer (including 0). B1 4! Seen either alone or in t – 4!, t an integer value > 24. M1 6! 4! , 1, 2 and 1, 2. 3! r u r u      = 96 A1 4 Question Answer Marks Guidance 7(c) Method 1 2s 4s 1,3,5 0 0 3 3C3 1 0 1 2 3C1  3C2 9 1 0 2 2C1  3C2 6 1 1 1 2C1  3C1  3C1 18 [Total 34 ways] M1 One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. A1 Two correct outcomes evaluated, accept unsimplified. M1 Four correct scenarios added. [Total number of selections = ] 8C3 [= 56] B1 Used as denominator of probability expression. [Probability =] 8 3 34 17 , 0.607 28 C       A1 Question Answer Marks Guidance 7(c) Method 2 Combinations of 3 numbers 1,2,3 1C1  2C1  1C1 2 1,2,4 1C1  2C1  3C1 6 1,2,5 1C1  2C1  1C1 2 1,3,4 1C1  1C1  3C1 3 1,3,5 1C1  1C1  1C1 1 1,4,5 1C1  3C1  1C1 3 2,3,4 2C1  1C1  3C1 6 2,3,5 2C1  1C1  1C1 2 2,4,5 2C1  3C1  1C1 6 3,4,5 1C1  3C1  1C1 3 (M1) One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. (A1) Five correct outcomes evaluated, accept unsimplified. (M1) Ten correct scenarios added. [Total 34 ways] [Total number of selections = ] 8C3 [= 56] (B1) 8C3 or 56 as denominator of probability expression. [Probability =] 8 3 34 17 , 0.607 28 C       (A1) Question Answer Marks Guidance 7(c) Method 3 1,2,3 1 2 1 3! 8 7 6    12 336 1,2,4 1 2 3 3! 8 7 6    36 336 1,2,5 1 2 1 3! 8 7 6    12 336 1,3,4 1 1 3 3! 8 7 6    18 336 1,3,5 1 1 1 3! 8 7 6    6 336 1,4,5 1 3 1 3! 8 7 6    18 336 2,3,4 2 1 3 3! 8 7 6    36 336 2,3,5 2 1 1 3! 8 7 6    12 336 2,4,5 2 3 1 3! 8 7 6    36 336 3,4,5 2 3 1 3! 8 7 6    18 336 (M1) One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4. (A1) Five correct outcomes evaluated, accept unsimplified. (M1) Ten correct scenarios added. Question Answer Marks Guidance 7(c) (B1) 336 or 8×7×6 seen as a denominator. [Probability ] 17 , 0.607 28  (A1) Method 4 444 3 2 1 8 7 6   6 336 445 3 2 1 3 8 7 6    18 336 443 2 3 1 3 8 7 6    18 336 442 2 3 1 3 8 7 6    36 336 441 2 3 1 3 8 7 6    18 336 225 2 3 1 3 8 7 6    6 336 224 2 3 1 3 8 7 6    18 336 223 2 3 1 3 8 7 6    6 336 221 2 3 1 3 8 7 6    6 336 (M1) 1-1 correct calculation, unsimplified for an identified scenario not containing three 4s. (A1) Five correct probabilities evaluated, accept unsimplified. 7(c) (M1) Nine correct scenarios subtracted. Question Answer Marks Guidance (B1) 336 or 8  7  6 seen as a denominator. [Probability] 132 204 1 , , 0.607 336 336  (A1) 5

Question 39

(a) How many different arrangements are there of the 10 letters in the word REGENERATE? [1] ...............................................…

Accepted Answer

7(a) 10! 2!4!        75600 1 7(b) 4! × 3! M1 4! SOI in all terms leading to final answer. Allow 24 if 4! = 24 is seen. M1 Ignoring any values used to justify 4!. Either 3! SOI in expression leading to final answer, or at least 6 distinct scenarios identified and added in expression leading to final answer. Condone 3 distinct scenarios × 2. Ignore repeated scenarios. A1 4! × 3! Fully correct unsimplified expression leading to final answer. 144 B1 WWW 4 Question Answer Marks Guidance 7(c) Method 1: If denominator is from 7(a), no denominator or incorrect denominator [Numerator = Number of required arrangements =] 5! 5! 2 2! 4!   [ = 600] B1 5! 2! seen (arrangements of consonants). B1 5! 4! seen (arrangements of vowels). M1 5! 5! 2   r s , r = 1 or 2, s = 1, 4, 4! or 24. [Probability =] 600 75600 their their M1  600 their their a or 600. 75600 their = 1 126 , 0.00794 A1 Accept 600 75600 OE. Method 2: If denominator 10! [Numerator = Number of required arrangements =] 5! 5! 2   [ = 28800] (B1) 5! seen (arrangements of consonants). (B1) A second 5! seen (arrangements of vowels). (M1) 5! 5! ,k   k = 1 or 2. [Probability = ] 28800 10! their (M1) = 1 126 , 0.00794 (A1) Accept 600 75600 OE. Question Answer Marks Guidance 7(c) Method 3: Using probabilities 5 5 4 4 3 3 2 2 1 1 2 10 9 8 7 6 5 4 3 2 1          (B1) 5 4 3 2 1     a b c d e seen. 10 ≥ a > b > c > d > e ≥ 1 (arrangements of consonants). (B1) A second 5 4 3 2 1     f g h i j seen. 10 ≥ f > g > h > i > j ≥ 1 (arrangements of vowels). (M1) 5 4 3 2 1 5 4 3 2 1 k a b c d e f g h i j           k = 1 or 2. (M1) 5 4 3 2 1 5 4 3 2 1. 10 9 8 7 6 5 4 3 2 1 their   = 1 126 , 0.00794 (A1) Accept 28800 362800 OE. 5

Question 40

(a) How many different arrangements are there of the 9 letters in the word RECORDERS? [1] .................................................…

Accepted Answer

6(a) 9! 30240 2!3!        1 6(b) Method 1: Number of arrangements with E at each end – Number of arrangements with E at each end and the three Rs together 7! 5! 3!   B1 7! 3! e  , 7P4 – e, e a positive integer. M1 5! ! f r  , f > 120, r = 1, 2 720 A1 If no marks scored SC B1 for 840 – 120 = 720. Method 2: Number of arrangements with E at each end and no Rs together + No of arrangements with E at each end and two Rs together 5C3 × 4! + 4C1 × 5! or 3 5P 4! 3!  + 5P2 4!  (B1) One of 5C3 × 4!, 3 5P 4! , 3!  4C1 × 5! or 5P2 4!  seen. (M1) a × 4! + b × 5! where a and b are integers between 1 and 10 inclusive, or c × 4! + d × 4! where c and d are integers between 1 and 20 inclusive. 240 + 480 = 720 (A1) 3 Question Answer Marks Guidance 6(c) Method 1 Group of 5 3 Rs 2 Es = 1 3 Rs 1 E = 2C1 × 4C1 = 8 3 Rs 0 Es = 4C2 = 6 Group of 4 3 Rs 1 E = 2C1 = 2 3 Rs 0 Es = 4C1 = 4 B1 Correct no of ways for two correct identified scenarios other than three Rs two Es. [Total =] 21 M1 No of ways for five correct identified scenarios added or correct. [Number of ways of splitting into the two groups =] 9C5 (= 126) seen as a denominator M1 Accept evaluated, accept 9C4. Probability = 21 1 126 6  (0.167) A1 Question Answer Marks Guidance 6(c) Method 2 3Rs in Group of 5 = 6C2 = 15 3Rs in Group of 4 = 6C1 = 6 (B1) One correct case evaluated accurately and linked with correct scenario. [Total =] 21 (M1) No of ways for two correct scenarios added or correct. [Number of ways of splitting into the two groups =] 9C5 (= 126) seen as a denominator (M1) Accept evaluated, accept 9C4. Probability = 21 1 126 6  (0.167) (A1) Method 3: Considering the possible positions of R within the groups 3Rs in Group of 5 5 4 3 9 8 7   = 15 126 3Rs in Group of 4 4 3 2 9 8 7   6 126  (B1) For one correct product unsimplified and linked with correct scenario. (M1) For second correct product. 15 126 + 6 126 (M1) For adding probabilities of two correct scenarios or correct. Probability = 21 1 126 6  (0.167) (A1) Question Answer Marks Guidance 6(c) Method 4: Probability method Group of 5 2Es 3 2 1 2 1 5! 1 9 8 7 6 5 3!2! 126       1E 4 3 2 1 2 5! 8 9 8 7 6 5 3! 126       0E 3 2 1 4 3 5! 6 9 8 7 6 5 3!2! 126       Group of 4 3 2 1 6 4! 6 9 8 7 6 3! 126      (B1) Two correct probabilities linked with correct scenarios, accept unsimplified. (M1) Four probabilities with denominators including a factor of 9  8  7  6  n, where n is 1 or 5. 1 8 6 6 126    (M1) Probabilities of four correct scenarios added or correct. Probability = 21 1 126 6  (0.167) (A1) 4

Question 41

(a) How many different arrangements are there of the 9 letters in the word INTELLECT in which the two Ts are together? [2] ................…

Accepted Answer

7(a) 8! M1 k ! k = 7 or 8, m = 1, 2. 2!2! 2! m ! = 10080 A1 7(b) Method 1 Number of ways with no restriction on Es – ways with Es together 7! 6! M1 7! − – r, r integer > 1. 2!2! 2! 2!2! [= 1260 – 360] M1 6! s − , s integer > 360. 2! = 900 A1 Method 2 T ^ ^ ^ ^ ^ T with Es inserted in gaps 5! 6  5 5! 6 M1 6  5 6  or  C 2 t  or t  C2 , t an integer > 1. 2! 2 2! 2 [= 60 × 15] M1 5!  u , u an integer > 1. 2! =900 A1 3 7(c) Method 1 – addition T E _ _ = 2C1 2C1  5C2 = 40 B1 Either identified or correct unsimplified expression, either alone or in an addition. T E E _ = 2C1 2C2  5C1 = 10 B1 Either identified or correct unsimplified expression, either alone or in an addition. M1 a ( 40 + 10 ) , a an integer < 126. Probability 9 9 C 4 C 4 Denominator value must be seen as 9 C 4 somewhere.  50  A1 39.68 ⩽ percentage ⩽ 39.7. Percentage =  100 = 39.7%    126  Method 2 – subtraction (total arrangements with 1 T – number of arrangements with 1T 0 E) T ^ ^ ^ = 2C1  7C3 = 70 B1 Either identified or correct unsimplified expression, either alone or in a subtraction. T * * * = 2C1  5C3 = 20 B1 Either identified or correct unsimplified expression, either alone or in a subtraction. M1 a ( 70 − 20 ) , a an integer < 126. Probability 9 9 C 4 C 4 Denominator value must be seen as 9 C 4 somewhere.  50  A1 39.68 ⩽ percentage ⩽ 39.7. Percentage =  100 = 39.7%    126  4

Question 42

(a) Find the number of different arrangements of the 9 letters in the word HAPPINESS. [1] .................................................…

Accepted Answer

6(a)  9!  B1 = 90720    2!2!  1 6(b) Method 1 Total arrangements – arrangements with repeated letters at ends 9! 7! M1 7! −  2 a −  b a = their 6(a) or correct, b = 1,2. 2!2! 2! 2! M1 7! a −  2 a = their 6(a) or correct, c = 1,2. c ! 85680 A1 FT ft their 6(a) – 5040. Method 2 Adding no of different ways P and S at ends 2  7! = 10080 M1 Finding correct number of ways for one of these correctly identified scenarios. 7! P or S at one end only 4 5 = 50400 2! M1 Adding no of ways for 3 correctly identified 7! scenarios. Neither P nor S at an end 5 4 = 25200 2!2! Total 85680 A1 Method 3 7! M1 Finding correct number of ways for one of P at beginning 7  = 17640 these correctly identified scenarios. 2! 7! S at beginning 7  = 17640 M1 Adding no of ways for 3 correctly identified 2! scenarios. 8! Neither P nor S at beginning 5  = 50400 2!2! Total 85680 A1 3 6(c) Method 1 arrangements with PP between Ss { S P P S ^ ^ ^ ^ ^ } add arrangements with PP not between Ss { (S ^ ^ S) ^ P P ^ } 6!+ 5!5 4 M1 6! + d, d an integer ≥ 1, may be implied. 6!+ 5!5 4 M1 e + 5! f , e, f integers ≥ 1, may be implied. , e an integer ≥ 1, g = M1 e + g ! ( 5  4 or 5 P2 ) 4,5,6. [Total ]= 3120 A1 Method 2 - considers the 6 positions for S ^^S Positions 1 and 6 there are 5 5! ways M1 Identifying no of ways if S^^S is in position 1 or 6. Positions 2, 3, 4 and 5 there are 4 5! ways M1 Identifying no of ways if S^^S is in position 2, 3, 4 or 5. 2 +5 5! 4 4 5! M1 Adding no of ways for 6 scenarios ( or 26 × 5!). [Total] = 3120 A1 SC B1 for 3120 if any method marks are withheld. 4 6(d) Method 1 Either PP in the group of 5 or PP in the group of 4 5 C3 5 C2 M1 a  5 C 2 , a  5 C 3 , or 5 C 2 + 5 C 3 seen as a + , 9 C5 9 C5 numerator of one or two fractions where a is 1 or 2, no extra terms. 5 C3 + 5 C 2 M1 9 9 9 C 5 or C 4 seen (no addition, C5 multiplication) as a denominator of one or two fractions. 20 10 A1 Probability = , , 0.159 126 63 Method 2 Considering the positions of P and then S  5 4 4 3   5 4 4 3  M1 a × 5 × 4 × 4 × 3 seen as a numerator of a      +      fraction.  9 8 7 6   7 6 9 8  where a = 1 or 2.  5 4 4 3   5 4 4 3  M1 9 × 8 × 7 × 6 seen as a denominator of a      +      fraction.  9 8 7 6   7 6 9 8  10 A1 = 63 3

Question 43

Alissa has 10 different books from the series Squares and Circles. The books look similar except for their colour. There are 3 blue books, …

Accepted Answer

6(a)  10!  B1 CAO. = 151200    2!3!2!  1 6(b) Method 1 Total – Ys together 8! 7! M1 8! 7! − = − , a = 1, 2 b = 1, 2 c = 1, 3. 2!2! 2! 2!a !c ! a !b !c ! [10080 – 2520 =] 7560 A1 2 Method 2 3 Bs treated as a single unit ^ ^ ^ ^ BBB ^ and Ys inserted 6! 7 M1 6! 7 6! 7 6! 7 P2  2  C 2 or  P2 or 2! C d ! d ! d ! 2 or 6! 7  6 6! 6 + 5 + 4 + 3 + 2 + 1) 6! 7  6 or  or (  d ! e d ! 2! 2 d = 1, 2, 3 e = 1, 2. or 6! ( 6 + 5 + 4 + 3 + 2 + 1) 2! [360 × 21 =] 7560 A1 2 6(c) Method 1 8! *M1 8! 5 , b = 1,2, c = 1,3 and b ≠ c, d ≥ 1. 2!3! b !c !d DM1 Multiply by 5. 16 800 A1 Method 2 8 P4  5! *M1 8 P4 , b = 1,2, c = 1,3 and b ≠ c, d ≥ 1. 2!3! b !c !d DM1 Multiply by 5!. 16 800 A1 3 6(d) Method 1 Y R B _ 2 2 3C1  3C1 = 36 M1 One correct identified unsimplified expression Y R R B 2 1 3C1 = 6 (3C1 ≠ 3C2). Y R _ _ 2 2 3C2 = 12 Y R R _ 2 1 3C1 = 6 B1 Correct outcome/value for 2 clearly identified scenarios, accept unsimplified WWW. M1 Sum of 4 correct identified scenarios. 60 A1 Method 2 Y R _ _ 5C3 × 3C1 × 2C1 M1 5C3 seen with YR^^ identified. M1 5C3 × a, a = 2, 3, 6. B1 5C3 × 3C1 × 2C1 or 5C3 × 3 × 2. 60 A1 4

Question 44

(a) Find the number of different arrangements of the 8 letters in the word KANGAROO in which the two As are together and the two Os are not…

Accepted Answer

2(c) 2 4 M1 2 4  1  3 405 p (1 − p )  5 , 0 < p < 1, p ≠ 1 – p. =    5  4  4  4096 = 0.0989 A1 AWRT 2

Question 45

In a group of 20 musicians, there are 9 guitarists, 6 pianists and 5 drummers. 6 musicians are selected from these 20 to perform at a conce…

Accepted Answer

5(a) 3G 2P 1D: 9C3 6C2  5C1 = 6300 B1 Correct outcome/value for either the 1st or 2nd scenario clearly identified, accept un-simplified, WWW. 4G 1P 1D: 9C4 6C1  5C1 = 3780 M1 2 correct outcomes/values obtained, accept un-simplified 5G 0P 1D: 9C5 [6C0]  5C1 = 630 M1 Sum of 3 correct scenarios, may be identified by un-simplified expression. Condone 5C1 =5 and 6C1 = 6 Total: 10710 A1 CAO If one or both M marks not awarded, SCB1 for 10710 WWW. 4 5(b) Ways of selecting 1st band: 9C3 6C1  5C1 = 2520 M1 9C3 6C1  5C1 or 9C3 × 6 × 5 seen, Ways of selecting 2nd band: 6C3 5C1  4C1 = 400 condone 3 or 3! . Ways of selecting 3rd band: 1 4C1  3C1 = 12 M1 their 2520 × their 400 × their 12 seen, accept un-simplified, [Total number of ways =] 2520 400  12 = condone 3 or 3! . 12096000 A1 Condone 12100000. If one or both M marks not awarded, SCB1 for 12096000 (CAO) WWW. 3

Question 46

A set of friends consists of 7 men and 4 women. Three of the men are brothers: Ali, Ben and Charlie. (a) Find the number of different arran…

Accepted Answer

7(b) 7! × 4! × 2 M1 7! × r, where r is an integer, r > 1 or 4! × s, where s is an integer, s > 1. M1 7! 4!  t, where t is an integer, t ⩾ 1, t = 1 can be implied. 241920 A1 CAO. 3 7(c) 11C6  5C3  [2C2] M1 11Cv × 11-vCu × [11-v-uCw], u, v, w = 6, 3, 2 u ≠ v ≠ w. = 4620 A1 2 7(d) Method 1 Ali, Ben and Charlie must be in the group of 6 or the group of 3. group of 6: 8C3  5C3  (2C2) = 560 B1 560 seen, accept un-simplified. group of 3: 8C6 ( 2C2  3C3 ) = 28 B1 28 seen accept un-simplified. 560 + 28 M1 their ( 560 + 28 ) their ( 560 + 28 ) Probability they are in same group = or 4620 their (c) 4620 588 7 A1 = , , 0.127 4620 55 Method 2 Ali, Ben and Charlie must be in the group of 6 or the group of 3. 6 5 4 3 2 1 B1 4 Group of 6:    8C3 , 0.1212… seen, accept un-simplified. 11 10 9 8 7 6 33 1 4 =  56 = 462 33 3 2 1 1 B1 1 Group of 3:   = , 0.00606 ( 06 ) seen, accept un-simplified. 11 10 9 165 165 4 1 M1 4 1 Probability they are in same group = + their + their . 33 165 33 165 7 A1 = ,0.127 55 4

Question 47

A darts club has 12 members made up of 7 men and 5 women. Every Monday, a team of 4 is chosen at random to represent the club in a competit…

Accepted Answer

6(b) 12C4 × 8C4 [× 4C4] M1 jC4 × k , j = 12, 8 k a positive integer > 1. 34650 A1 SCM1 for 12C4 × 8C4[ × 4C4] 3! SCA1 for 5775. 2 6(c) Method 1 – summing no of ways with at least one man and one woman in each team 3M 1W + 3M 1W + 1M 3W M1 (7Cm × 5C4-m ) 1 ⩽ m ⩽ 3 seen multiplied by at least 3M 1W + 2M 2W + 2M 2W one other Combination in form nCr. 3! (7C3 × 5C1 ) × ( 4C3 × 4C1 ) [× (1C1 × 3C3)] × M1 No of ways for two correctly identified scenarios, or 2! correct, added, no incorrect. = 175 ×16 × 3= 8400 3! (7C3 × 5C1 ) × ( 4C2 × 4C2 ) [×(2C2 × 2C2 )] × 2! = 175 × 36 ×3 = 18900 27300 A1 SC A1 for 4550. SC B1 for 9100 if only one M1 has been awarded. Method 2 – subtracting ways with only men/women in a team from total 4M 0W + 3M 1W + 0M 4W M1 pC4 where p = 7, 6, 5 or 4 4M 0W + 2M 2W + 1M 3W seen multiplied by at least 1 other Combinations in form nCr, r  0, n  r . 7C4 [× 5C0] × 3C3 × 5C1 × 3! M1 No of ways for two correctly identified scenarios = 175 x 6 = 1050 added (or correct) and subtracted from 34650 or their (b). 7C4 [× 5C0] × 3C2 × 5C2 × 3! = 1050 × 6 = 6300 34650 – (1050 +6300) 27300 A1 SC A1 for 4550. 6(c) Method 3 – subtracting ways with only men/women in a team from total M1 (7C4 × 8C4) or (5C4 × 8C4) seen. 3! all male team 7C4 8C4  = 7350 M1 Subtracting (all male + all female – overlap) correctly 2! identified or correct from 34650 or their (b). 3! all female team 5C4 8C4  = 1050 2! 3! all male AND all female = 7C4 5C4  = 1050 1! 34650 – (7350+1050-1050) 27300 A1 SCA1 for 4550. Method 4 – subtracting ways with only men in a team as this includes the way with only women 3! M1 (7C4 × 8C4) seen. all male team 7C4 8C4  = 7350 2! M1 Subtracting from 34650 or their (b). 34650 – 7350 27300 A1 SCA1 for 4550. 3 6(d) Method 1 G _ _ H _ _ _ M1 5! × n , n = 2,4,8. 5! × 2 × 4 960 A1 Method 2 5P2 × 2! × 4! or 5C2 × 2 × 2! × 4! M1 5P2 × n or 5C2 × 2 × n where n = 2!, 4! or 2! ×4! 960 A1 2

Question 48

(a) How many different arrangements are there of the 10 letters in the word SEYCHELLES? [1] ...............................................…

Accepted Answer

7(a)  10!  B1 CAO. = 151200    2!2!3!  1 7(b) Method 1[SC_ S _ _ _ _ _ _] 7! M1 7! 2 7 2!3!k 1  k , a positive integer. 2!3! M1 Integer 7 . = 5880 A1 3 Method 2 Considering each case separately [SC-S, S-CS] 6! M1 6! 6! 6! With Y =2 7 840 , and seen. 2!3! 2!3!l 2!2!m 3!n 6! With H =2 7 840 1  l , m, n , positive integers. 2!3! 6! With E 2 7 = 2520 2!2! 6! With L 2 7 = 1680 3! 840 + 840 + 2520 + 1680 M1 Summing 4 correct or correctly identified scenarios, oe = 5880 A1 7(b) Method 3 Considering each case separately [SC-S, S-CS] 7! M1 7! 7! 7! SCYS 2!3! =2 840 2!3!l , 2!2!m and 2!n seen 7! SCHS =2 840 1  l , m, n , positive integers 2!3! 7! SCES  2 = 2520 2!2! 7! SCLS  2 = 1680 3! 840 + 840 + 2520 + 1680 M1 Summing 4 correct or correctly identified scenarios, OE. = 5880 A1 3 7(c) Method 1 Total arrangements with Ss at ends – arrangements with Ss at ends and Es together 8! 6! M1 8! − 2!3! 2! 2!3!− q , 1  q  3360 . [= 3360 – 360] M1 m − 6!, m  360 . 2! = 3000 A1 3 7(d) Method 1 [Number of ways with 3 Es =] 7C2 (= 21) B1 7C2 seen with no addition, subtraction, multiplication. [Total number of ways is] 10C5 (= 252) M1 Seen. 21 1 A1 21 1 [Probability =] , If M mark not awarded, SCB1 for , WWW. 252 12 252 12 Method 2 3 2 3 B1 3 2 3    5C2   . 11 6 11 11 6 11 M1 5C2  k , 0  k  1 . Accept 5C3  k , 0  k  1 . 21 1 A1 21 1 [Probability =] , If M mark not awarded, SCB1 for , WWW. 252 12 252 12 3

Question 49

(a) Find the number of different arrangements of the 10 letters in the word ZOOLOGICAL in which the three Os are together and the two Ls ar…

Accepted Answer

7(c) Method 1 1 – P(2 letters the same)  3 2 2 1  M1 3 2 2 1 1 −   +    +  seen.  10 9 10 9  10 9 10 9 M1  3 2 2 1  1 −   +   d = 9 or 10.  d d d d   6 3 2 1 2 1  Accept 1 −  or or or + or  .  90 45 30 15 90 45  A1 = 41, 0.911 If one or more M mark not awarded, SCB1 for 41, 0.911 WWW. 45 45 Method 2 P(O O) + P(L L) + P( OL) 3 7 2 8 5 9  +  +  10 9 10 9 10 9 M1 3 7 2 8 5 9  +  +  , d = 9 or 10. d d d d d d Accept   21 7   16 8   45 9 1     or  +  or  +  or or   .   90 30   90 45   90 18 2   A1 = 41, 0.911 If one or more M mark not awarded, SCB1 for 41, 0.911 WWW. 45 45 7(c) Method 3 Combination approach using OO and LL 3 2 M1 3C2 + 2C2 seen. C 2 + C 2 [Probability =] 1 − 10 C 2 M1 f 1 − , 1 ⩽ f < 45. 10 C 2 A1 = 41, 0.911 If one or more M mark not awarded, SCB1 for 41, 0.911 WWW. 45 45 Method 4 Combinations with scenarios OL, OL, OL, OL M1 3C1 × 2C1 + 3C1 × 5C1 + 5C1 × 2C1 + 5C2 seen O and L 3C1 × 2C1 [6] M1 g ,1  g  45 10 O and not L 3C1 × 5C1 [15] C 2 Not O and L 5C1 × 2C1 [10] Not O and not L 5C2 [10] 6 + 15 + 10 + 10 [Probability = ] 10 C 2 A1 = 41, 0.911 If one or more M mark not awarded, SCB1 for 41, 0.911 WWW. 45 45 3

Question 50

The Splash Club has 26 members, of whom 16 are swimmers and 10 are divers. No member is both a swimmer and a diver. The club committee cons…

Accepted Answer

2 B1 Expression of the form 16Cx  10Cy with x+ y = 6 linked to a Scenario S(16) D(10) correct identified scenario. SSSSDD 4 2 16C4  10C2 [81900] SSSDDD 3 3 16C3  10C3 [67200] SSDDDD 2 4 16C2  10C4 [25200] M1 Two identified outcomes evaluated accurately, accept un- simplified. Identification can be implied by un-simplified expression. Condone consistent use of permutations. M1 Sum of their values of 3 correct identified scenarios, no incorrect/repeated scenarios. Identification can be implied by un-simplified expression. Total = 174 300 A1 If either or both Ms not awarded, SCB1 for 174 300 WWW. 4

Question 51

(a) Find the number of different arrangements of the 9 letters in the word DAFFODILS in which there is a D at each end and the two Fs are n…

Accepted Answer

3(a) Method 1 Total arrangements with Ds at ends – arrangements with Ds at ends and Fs together 7! M1 7! − 6! seen alone, not multiplied or divided in a calculation nor 2! 2! subtracted from another value. [= 2520 – 720] M1 l ! − 6!, l = 7, 8 or 9, m = 1 or 2 m ! = 1800 A1 Method 2 D ^ ^ ^ ^ ^ D, Fs inserted not next to each other 6  5 M1 5!  a, 1 a 30 1 can be implied, a is an integer. No other 5!  2 terms added or subtracted. M1 6  5 6 P 2 b !  or b ! 6C2 orb ! . 2 2! Where b = 5, 6 or 7 . = 1800 A1 3 3(b) Method 1 [Number of outcomes with 4 letters between Ds (D ^ ^ ^ ^ D ^ ^ ^)] M1 Accept 7! 7!  4 [= 10080]  4 or 10080 alone or as the numerator or denominator of a 2! 2 fraction. 9! M1 Accept 90720 as denominator in a fraction. [Total number of arrangements =] [= 90720] seen as 2!2! denominator in a fraction 1 A1 WWW.  Probability = 10080 , , 0.111 90720 9 Method 2 [Total number of outcomes with 4 letters between Ds (D1 ^ ^ ^ ^ D2 ^ M1 Accept 7!  4  2! or 40320 alone or as the numerator or ^ ^)] denominator of a fraction. 7!  4  2! [= 40320] [Total number of arrangements of D1AF1F2OD2ILS =] M1 Accept 362880 as denominator in a fraction. 9! [=362880] seen as denominator in a fraction. 1 A1 WWW.  Probability = 40320 , , 0.111 362880 9 3(b) Method 3 [When 2 F’s are not part of the 4 letters between the D’s] M1 Accept 10080 alone or as the numerator or denominator of a fraction. 5P4  4! 5!  4! or [= 1440] 2! 2! [When 1F is part of the 4 letters between the D’s] 5C3  4!  4! [=5760] [When 2F’s are part of the 4 letters between the D’s] 5C2  4!  4! [=2880] 2! [1440 + 5760 + 2880] = 10080 9! M1 Accept 90720 as denominator in a fraction. [Total number of arrangements =] [= 90720] seen as 2!2! denominator in a fraction 1 A1 WWW.  Probability = 10080 , , 0.111 90720 9 3

Question 52

In a group of 25 athletes, there are 8 sprinters, 5 hurdlers and 12 throwers. (a) Find the number of different ways in which a team of 6 at…

Accepted Answer

5(a) 6S 8C6 (= 28) M1 One product using 2 or 3 combinations 8Cx  5Cy  12Cz with x + 5S 1H 8C5  5C1 (= 280) y + z = 6 (one of y or z may be 0) and linked to a correct identified 5S 1T 8C5  12C1 (= 672) scenario. 4S 2H 8C4  5C2 (= 700) 4S 1H 1T 8C4  5C1  12C1 (= 4200) B1 2 correct identified outcomes, accept un-simplified. 3S 2H 1T 8C3  5C2  12C1 (= 6720) Identification can be implied by un-simplified expressions. 2nd and 3rd can be combined as 8C5  17C1 and this counts as 2 M1 Add values of 6 correct scenarios, no incorrect/repeated scenarios, scenarios for the second M1 no extra scenarios. Identification can be implied by un-simplified expressions. Total = 12600 A1 If either or both Ms not awarded, SCB1 for 12600 WWW. 4 5(b) Method 1 Total arrangements – arrangements with 3 hurdlers together [With 3 hurdlers not together ] 8! − (6!×3!) B1 6!  3! seen as a term added or subtracted. M1 8! − ( 6!  k ) , k  1 , k is an integer. = 36000 A1 Method 2 Summing no of ways with 2 hurdlers together or all separate 6C 3  3!  5! + 6C 2 2 3!  5! B1 6Cx × 5! or 6Px × 5! seen, x = 2 or 3 seen.  = 14400 + 21600  M1 ( 6C 3  5! + 6C 2  2  5! ) k , or 6 P3 +5! 6 P2  5! k k  1, k is an integer. = 36000 A1 3

Question	Answer	Marks	From
1	see sheet	3	9709/52 Feb/March 2020
2	see sheet	6	9709/52 Feb/March 2020
3	see sheet	6	9709/51 May/June 2020
4	see sheet	4	9709/51 May/June 2020
5	see sheet	9	9709/52 May/June 2020
6	see sheet	10	9709/53 May/June 2020
7	see sheet	11	9709/51 Oct/Nov 2020
8	see sheet	11	9709/52 Oct/Nov 2020
9	see sheet	6	9709/53 Oct/Nov 2020
10	see sheet	7	9709/53 Oct/Nov 2020
11	see sheet	10	9709/52 Feb/March 2021
12	see sheet	4	9709/51 May/June 2021
13	see sheet	8	9709/51 May/June 2021
14	see sheet	10	9709/52 May/June 2021
15	see sheet	10	9709/53 May/June 2021
16	see sheet	10	9709/51 Oct/Nov 2021
17	see sheet	5	9709/52 Oct/Nov 2021
18	see sheet	6	9709/52 Oct/Nov 2021
19	see sheet	2	9709/53 Oct/Nov 2021
20	see sheet	8	9709/53 Oct/Nov 2021
21	see sheet	10	9709/52 Feb/March 2022
22	see sheet	6	9709/51 May/June 2022
23	see sheet	6	9709/51 May/June 2022
24	see sheet	10	9709/52 May/June 2022
25	see sheet	11	9709/53 May/June 2022
26	see sheet	10	9709/51 Oct/Nov 2022
27	see sheet	10	9709/52 Oct/Nov 2022
28	see sheet	10	9709/53 Oct/Nov 2022
29	see sheet	12	9709/52 Feb/March 2023
30	see sheet	5	9709/51 May/June 2023
31	see sheet	6	9709/51 May/June 2023
32	see sheet	10	9709/52 May/June 2023
33	see sheet	9	9709/53 May/June 2023
34	see sheet	10	9709/51 Oct/Nov 2023
35	see sheet	10	9709/52 Oct/Nov 2023
36	see sheet	10	9709/53 Oct/Nov 2023
37	see sheet	10	9709/52 Feb/March 2024
38	see sheet	10	9709/51 May/June 2024
39	see sheet	10	9709/52 May/June 2024
40	see sheet	8	9709/53 May/June 2024
41	see sheet	9	9709/51 Oct/Nov 2024
42	see sheet	11	9709/53 Oct/Nov 2024
43	see sheet	10	9709/52 Feb/March 2025
44	see sheet	7	9709/51 May/June 2025
45	see sheet	7	9709/51 May/June 2025
46	see sheet	12	9709/53 May/June 2025
47	see sheet	10	9709/55 May/June 2025
48	see sheet	10	9709/51 Oct/Nov 2025
49	see sheet	10	9709/52 Oct/Nov 2025
50	see sheet	4	9709/53 Oct/Nov 2025
51	see sheet	6	9709/53 Oct/Nov 2025
52	see sheet	10	9709/55 Oct/Nov 2025

Questions86 pages

Mark scheme52 answers

Another paper, or another topic

Questions as text

Q1 · The 40 members of a club include Ranuf and Saed 9709/52 Feb/March 2020

Q2 · Richard has 3 blue candles, 2 red candles and 6 green candles 9709/52 Feb/March 2020

Q3 · Find the number of diﬀerent arrangements that can be made from the 9 letters of the word… 9709/51 May/June 2020

Q4 · In a music competition, there are 8 pianists, 4 guitarists and 6 violinists 9709/51 May/June 2020

Q5 · Find the number of diﬀerent ways in which the 10 letters of the word SUMMERTIME can be… 9709/52 May/June 2020

Q6 · Find the number of diﬀerent possible arrangements of the 9 letters in the word CELESTIAL 9709/53 May/June 2020

Q7 · Find the number of diﬀerent ways in which the 10 letters of the word SHOPKEEPER can be… 9709/51 Oct/Nov 2020

Q8 · Mr and Mrs Ahmed with their two children, and Mr and Mrs Baker with their three children… 9709/52 Oct/Nov 2020

Q9 · A committee of 6 people is to be chosen from 9 women and 5 men 9709/53 Oct/Nov 2020

Q10 · The 8 letters in the word RESERVED are arranged in a random order 9709/53 Oct/Nov 2020

Q11 · Find the total number of diﬀerent arrangements of the 11 letters in the word CATERPILLAR 9709/52 Feb/March 2021

Q12 · A bag contains 12 marbles, each of a diﬀerent size 9709/51 May/June 2021

Q13 · How many diﬀerent arrangements are there of the 8 letters in the word RELEASED? 9709/51 May/June 2021

Q14 · Find the total number of diﬀerent arrangements of the 8 letters in the word TOMORROW 9709/52 May/June 2021

Q15 · How many diﬀerent arrangements are there of the 11 letters in the word REQUIREMENT? 9709/53 May/June 2021

Q16 · Raman and Sanjay are members of a quiz team which has 9 members in total 9709/51 Oct/Nov 2021

Q17 · A group of 6 people is to be chosen from 4 men and 11 women 9709/52 Oct/Nov 2021

Q18 · In how many diﬀerent ways can the 9 letters of the word TELESCOPE be arranged? 9709/52 Oct/Nov 2021

Q19 · The 26 members of the local sports club include Mr and Mrs Khan and their son Abad 9709/53 Oct/Nov 2021

Q20 · A security code consists of 2 letters followed by a 4-digit number 9709/53 Oct/Nov 2021

Q21 · A group of 12 people consists of 3 boys, 4 girls and 5 adults 9709/52 Feb/March 2022

Q22 · Find the number of diﬀerent arrangements of the 8 letters in the word DECEIVED in which… 9709/51 May/June 2022

Q23 · There are 6 men and 8 women in a Book Club 9709/51 May/June 2022

Q24 · Find the number of diﬀerent arrangements of the 9 letters in the word CROCODILE 9709/52 May/June 2022

Q25 · A group of 15 friends visit an adventure park 9709/53 May/June 2022

Q26 · A Social Club has 15 members, of whom 8 are men and 7 are women 9709/51 Oct/Nov 2022

Q27 · Find the number of diﬀerent arrangements of the 9 letters in the word ALLIGATOR in which… 9709/52 Oct/Nov 2022

Q28 · Find the number of diﬀerent arrangements of the 9 letters in the word ACTIVATED 9709/53 Oct/Nov 2022

Q29 · Find the number of diﬀerent arrangements of the 9 letters in the word DELIVERED in which… 9709/52 Feb/March 2023

Q30 · Find the number of ways in which a committee of 6 people can be chosen from 6 men and 8… 9709/51 May/June 2023

Q31 · Find the number of diﬀerent arrangements of the 8 letters in the word COCOONED 9709/51 May/June 2023

Q32 · In a group of 25 people there are 6 swimmers, 8 cyclists and 11 runners 9709/52 May/June 2023

Q33 · Find the number of diﬀerent arrangements of the 10 letters in the word CASABLANCA in… 9709/53 May/June 2023

Q34 · Table X Table Y In a restaurant, the tables are rectangular 9709/51 Oct/Nov 2023

Q35 · Find the number of diﬀerent arrangements of the 9 letters in the word ANDROMEDA in which… 9709/52 Oct/Nov 2023

Q36 · Jai and his wife Kaz are having a party 9709/53 Oct/Nov 2023

Q37 · A new village social club has 10 members of whom 6 are men and 4 are women 9709/52 Feb/March 2024

Q38 · The eight digits 1, 2, 2, 3, 4, 4, 4, 5 are arranged in a line 9709/51 May/June 2024

Q39 · How many different arrangements are there of the 10 letters in the word REGENERATE? 9709/52 May/June 2024

Q40 · How many different arrangements are there of the 9 letters in the word RECORDERS? 9709/53 May/June 2024

Q41 · How many different arrangements are there of the 9 letters in the word INTELLECT in which… 9709/51 Oct/Nov 2024

Q42 · Find the number of different arrangements of the 9 letters in the word HAPPINESS 9709/53 Oct/Nov 2024

Q43 · Alissa has 10 different books from the series Squares and Circles 9709/52 Feb/March 2025

Q44 · Find the number of different arrangements of the 8 letters in the word KANGAROO in which… 9709/51 May/June 2025

Q45 · In a group of 20 musicians, there are 9 guitarists, 6 pianists and 5 drummers 9709/51 May/June 2025

Q46 · A set of friends consists of 7 men and 4 women 9709/53 May/June 2025

Q47 · A darts club has 12 members made up of 7 men and 5 women 9709/55 May/June 2025

Q48 · How many different arrangements are there of the 10 letters in the word SEYCHELLES? 9709/51 Oct/Nov 2025

Q49 · Find the number of different arrangements of the 10 letters in the word ZOOLOGICAL in… 9709/52 Oct/Nov 2025

Q50 · The Splash Club has 26 members, of whom 16 are swimmers and 10 are divers 9709/53 Oct/Nov 2025

Q51 · Find the number of different arrangements of the 9 letters in the word DAFFODILS in which… 9709/53 Oct/Nov 2025

Q52 · In a group of 25 athletes, there are 8 sprinters, 5 hurdlers and 12 throwers 9709/55 Oct/Nov 2025