Solve problems on arrangement and selection — Paper 2 · IGCSE Mathematics - Additional 0606

5 (a) How many 5-digit numbers are there that have 5 different digits and are divisible by 5? [3] (b) A committee of 8 people is to be selected from 9 men and 5 women. Find the number of different committees that can be selected if the committee must have at least 4 women. [3]

6 marks

Mark scheme: 5(a) ( 9 × 8 × 7 × 6 × 1) + ( 8 × 8 × 7 × 6 × 1 ) soi M2 M1 for one correct product of the sum 5712 A1 5(b) 9C4 × 5C4 + 9C3 × 5C5 oe M2 M1 for one correct product of the sum [630 + 84 = ] 714 A1

This question in 0606/23 May/June 2017

5 Naomi is going on holiday and intends to read 4 books during her time away. She selects these books from 5 mystery, 3 crime and 2 romance books. Find the number of ways in which she can make her selection in each of the following cases. (i) There are no restrictions. [1] (ii) She selects at least 2 mystery books. [3] (iii) She selects at least 1 book of each type. [3]

7 marks

Mark scheme: 5(i) 10C 4 = 210 B1 5(ii) 2 Mystery 2 others = 5C2 × 5C2 =100 B3 B1 for one combination, 5 5 unsimplifiied 3 Mystery 1 other = C3 × C1 = 50 B1 for second combination, 4 Mystery = 5C 4 =5 unsimplifiied B1 for third combination, Total 155 unsimplifiied and total Alternative Method All – 0 Mystery – 1 Mystery B1 All minus 0 or 1 or both = 210 − 5C 4 − 5C1 × 5C3 B1 B1dep 1Mystery and 0 mystery unsimplified = 210 − 5 −×5 10 = 155 B1 B1dep final answer 5(iii) 2M1C1R = 5C 2 × 3C1 × 2C1 = 60 B3 B1 for one combination, 5 3 2 unsimplifiied 1M2C1R = C1 × C 2 × C1 = 30 B1 for second combination, = 5C1 × 3C1 × 2C 2 unsimplifiied 1M1C2R B1 for third combination, = 15 unsimplifiied and total Total 105

This question in 0606/22 Oct/Nov 2017

3 A 7-character password is to be selected from the 12 characters shown in the table. Each character may be used only once. Characters Upper-case letters A B C D Lower-case letters e f g h Digits 1 2 3 4 Find the number of different passwords (i) if there are no restrictions, [1] (ii) that start with a digit, [1] (iii) that contain 4 upper-case letters and 3 lower-case letters such that all the upper-case letters are together and all the lower-case letters are together. [3]

5 marks

Mark scheme: 3(i) ( 12 P7 = ) 3991680 B1 3(ii) 11 B1 (4 × 6P = ) 1 330 560 3(iii) 4! × 4! × 2 oe M2 M1 for 4! × 4! oe only or 4 P4 × 4 P3 oe only 1152 A1

This question in 0606/21 May/June 2018

3 A 7-character password is to be selected from the 12 characters shown in the table. Each character may be used only once. Characters Upper-case letters A B C D Lower-case letters e f g h Digits 1 2 3 4 Find the number of different passwords (i) if there are no restrictions, [1] (ii) that start with a digit, [1] (iii) that contain 4 upper-case letters and 3 lower-case letters such that all the upper-case letters are together and all the lower-case letters are together. [3]

5 marks

Mark scheme: 3(i) ( 12 P7 = ) 3991680 B1 3(ii) 11 B1 (4 × 6P = ) 1 330 560 3(iii) 4! × 4! × 2 oe M2 M1 for 4! × 4! oe only or 4 P4 × 4 P3 oe only 1152 A1

This question in 0606/23 May/June 2018

6 A 5-digit code is to be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. Each digit can be used once only in any code. Find how many codes can be formed if (i) the first digit of the code is 6 and the other four digits are odd, [2] (ii) each of the first three digits is even, [2] (iii) the first and last digits are prime. [2]

6 marks

Mark scheme: 6(i) 120 2 B2 5 × 4 × 3 × 2 or B1for pattern n ( n − 1)( n − 2)( n − 3) 6(ii) 720 2 B1 4 × 3 × 2 B1 dep × 6 × 5 = 720 6(iii) 2520 2 B1 4 ×…×…×…× 3 B1 Dep × 7 × 6 × 5 = 2520

This question in 0606/21 Oct/Nov 2018

11 R H x 15 9 x x – 2 N There are 70 girls in a year group at a school. The Venn diagram gives some information about the numbers of these girls who play rounders (R), hockey (H) and netball (N). n(R) = 28 n(H) = 38 n(N) = 35. Find the value of x and hence the number of girls who play netball only. [6]

6 marks

Mark scheme: B1 ′11 n ( ( R ∩ H ) ∩ N ) = 14 − x B1 ′ n ( ( R ∩ N ) ∩ H ) = 5 n( N ∩ (R ∪ H )') = 21 − x B1 x + 9 + x + 15 + 14 − x + 5 + 21 − x + x − 2 M1 correctly form equation in x and = 70 attempt to solve x = 8 A1 n( N ∩ (R ∪ H )') = 13 A1

This question in 0606/21 Oct/Nov 2018

6 (a) A 5-character code is to be formed from the 13 characters shown below. Each character may be used once only in any code. Letters : A, B, C, D, E, F Numbers: 1, 2, 3, 4, 5, 6, 7 Find the number of different codes in which no two letters follow each other and no two numbers follow each other. [3] (b) A netball team of 7 players is to be chosen from 10 girls. 3 of these 10 girls are sisters. Find the number of different ways the team can be chosen if the team does not contain all 3 sisters. [3]

6 marks

Mark scheme: 6(a) Number first B1 = 7 × 6 × 5 × 6 × 5 or 7 P3 × 6 P2 or 6300 Letter first B1 = 6 × 5 × 4 × 7 × 6 or 6 P3 × 7 P2 or 5040 6300 + 5040 = 11 340 B1 6(b) With 2 sisters = 7 C5 × 3C 2 = 63 3 B1 One combination evaluated 7 3 B1Another combination With 1 sister = C6 × C1 = 21 evaluated With no sister = 7 C7 = 1 and B1 Third combination and 85 Total 85 OR Total no of ways = 10 C7 = 120 B1 With 3 sisters = 7 C 4 = 35 B1 Without 3 sisters = 120 − 35 = 85 B1

This question in 0606/22 Oct/Nov 2018

7 A squad of 20 boys, which includes 2 sets of twins, is available for selection for a cricket team of 11 players. Calculate the number of different teams that can be selected if (i) there are no restrictions, [1] (ii) both sets of twins are selected, [2] (iii) one set of twins is selected but neither twin from the other set is selected, [2] (iv) exactly one twin from each set of twins is selected. [2]

7 marks

Mark scheme: 7(i) 167 960 1 7(ii) evidence of selecting from 16 M1 [16 C7 =] 11 440 A1 7(iii) 2 × n Cr with n = 16 or r = 9 M1   2 ×16 C 9 =  22880 A1 7(iv) 4 × n Cr with n = 16 or r = 9 M1   4 ×16 C 9 =  45760 A1

This question in 0606/23 Oct/Nov 2018

1 A band can play 25 different pieces of music. From these pieces of music, 8 are to be selected for a concert. (i) Find the number of different ways this can be done. [1] The 8 pieces of music are then arranged in order. (ii) Find the number of different arrangements possible. [1] The band has 15 members. Three members are chosen at random to be the treasurer, secretary and agent. (iii) Find the number of ways in which this can be done. [1]

3 marks

Mark scheme: Question Answer Marks Partial Marks 1(i) 1 081 575 B1 1(ii) 40 320 B1 1(iii) 2730 B1

This question in 0606/22 Feb/March 2019

9 (a) Eleven different television sets are to be displayed in a line in a large shop. (i) Find the number of different ways the televisions can be arranged. [1] Of these television sets, 6 are made by company A and 5 are made by company B. (ii) Find the number of different ways the televisions can be arranged so that no two sets made by company A are next to each other. [2] (b) A group of people is to be selected from 5 women and 3 men. (i) Calculate the number of different groups of 4 people that have exactly 3 women. [2] (ii) Calculate the number of different groups of at most 4 people where the number of women is the same as the number of men. [2]

7 marks

Mark scheme: 9(a)(i) 39 916 800 B1 9(a)(ii) 5! × 6! oe M1 86 400 A1 9(b)(i) 5C3 × 3C1 oe M1 30 A1 9(b)(ii) 5C 2 × 3C 2 + 5C1 × 3C1 oe M1 45 A1

This question in 0606/21 May/June 2019

4 (a) (i) Find how many different 5-digit numbers can be formed using five of the eight digits 1, 2, 3, 4, 5, 6, 7, 8 if each digit can be used once only. [2] (ii) Find how many of these 5-digit numbers are greater than 60 000. [2] (b) A team of 3 people is to be selected from 4 men and 5 women. Find the number of different teams that could be selected which include at least 2 women. [2]

6 marks

Mark scheme: 4(a)(i) 6720 B2 B1 for 8 × 7 × 6 × 5 × 4 or 8 5P 4(a)(ii) 2520 B2 B1for 3 × 7 × 6 × 5 × 4 or 3 P1 × 7 P4 4(b) 4C1 × 5C2 + 5C3 M1 50 A1

This question in 0606/23 May/June 2020

6 A 4-digit code is to be formed using 4 different numbers selected from 1, 2, 3, 4, 5, 6, 7, 8 and 9. Find how many different codes can be formed if (a) there are no restrictions, [1] (b) only prime numbers are used, [1] (c) two even numbers are followed by two odd numbers, [2] (d) the code forms an even number. [2]

6 marks

Mark scheme: 6(a) 3024 B1 6(b) 24 B1 6(c) 4P2 × 5P2 M1 4×3 × 5×4 240 no isw A1 6(d) 4P1 × 8P3 M1 4 × 8×7×6 1344 no isw A1

This question in 0606/23 Oct/Nov 2020

8 A photographer takes 12 different photographs. There are 3 of sunsets, 4 of oceans, and 5 of mountains. (a) The photographs are arranged in a line on a wall. (i) How many possible arrangements are there if there are no restrictions? [1] (ii) How many possible arrangements are there if the first photograph is of a sunset and the last photograph is of an ocean? [2] (iii) How many possible arrangements are there if all the photographs of mountains are next to each other? [2] (b) Three of the photographs are to be selected for a competition. (i) Find the number of different possible selections if no photograph of a sunset is chosen. [2] (ii) Find the number of different possible selections if one photograph of each type (sunset, ocean, mountain) is chosen. [2]

9 marks

Mark scheme: 8(a)(i) 479 001 600 oe B1 8(a)(ii) 3 × 10! × 4 oe M1 43 545 600 oe A1 8(a)(iii) 5! × 8 × 7! oe M1 4 838 400 oe A1 8(b)(i) 9 C 3 M1 84 A1 8(b)(ii) 3 C1 × 4 C1 × 5 C1 oe M1 60 A1

This question in 0606/22 Feb/March 2021

8 Marc chooses 5 people from 4 men, 4 women and 2 children. Find the number of ways that Marc can do this (a) if there are no restrictions, [1] (b) if at least 2 men are chosen, [3] (c) if at least 1 man, at least 1 woman and at least 1 child are chosen. [3]

7 marks

Mark scheme: 8(a) 252 B1 8(b) [2 men and 3 others = ] 120 M2 M1 for any two correct [3 men and 2 others = ] 60 [4 men and 1 other = ] 6 186 A1 Alternative method [0 men =] 6 (M1) [1 man and 4 others = ] 60 (their 252) – (6 + 60) (M1) 186 (A1) 8(c) M W C M2 for at least four out of five 1 3 1 32 correct values soi 1 2 2 24 or M1 for any two or three 2 2 1 72 correct values soi 2 1 2 24 3 1 1 32 184 A1 Alternative method [0 men =] 6 (M1) [0 women ] 6 [0 children] 56 (their 252) – (6 + 6 + 56) (M1) 184 (A1)

This question in 0606/21 Oct/Nov 2021

3 A group of students, 4 girls and 3 boys, stand in line. (a) Find the number of different ways the students can stand in line if there are no restrictions. [1] (b) Find the number of different ways the students can stand in line if the 3 boys are next to each other. [2] (c) Cam and Dea are 2 of the girls. Find the number of ways the students can stand in line if Cam and Dea are not next to each other. [2]

5 marks

Mark scheme: 3(a) [7! = ] 5040 B1 3(b) 3! × 5! oe M1 720 A1 3(c) 5040 – (2! × 6!) oe M1 3600 A1

This question in 0606/22 Feb/March 2022

6 (a) (i) A 5-digit number is to be formed from the seven digits 0, 1, 2, 3, 4, 5, 6. Each digit can be used at most once in any number and the number does not start with 0. Find the number of ways in which this can be done. [2] (ii) Find how many of these 5-digit numbers are even. [3] (b) A team of 7 people is to be selected from a group of 9 women and 6 men. Find the number of different teams that can be selected which include at least one man. [2] n n 1 3(c) (i) Show that C 3 + C 2 = ( n - n ) for n H 3. [5] 6 (ii) Hence solve the equation n C 3 n + C 2 = 4n where n H 3. [2]

14 marks

Mark scheme: 6(a)(i) 6 × 6 × 5 × 4 × 3 oe M1 2160 A1 6(a)(ii) Full correct calculation (360 + 900) M2 M1 for any correct product ‘how many end with 0’+ soi ‘how many end with 2, 4, 6’ oe Eg (6 × 5 × 4 × 3 × 1) or 360 (5 × 5 × 4 × 3 × 1) or 300 (5 × 5 × 4 × 3 × 3) or 900 1260 cao A1 6(b) 15C7  9C7 M1 6399 A1 6(c)(i) n ! n ! B2 B1 for either expression  correct ( n  3)!3! ( n  2)!2! n ( n  1)( n  2) n ( n  1) M2 M1 for either expression  correct 6 2 n ( n  1)( n  2  3) A1 6 n 3  3n 2  2 n 3n 2  3n or  6 6 1 3 leading to ( n  n ) 6 6(c)(ii) n ( n 2  25)  0 oe M1 n = 5 as the only solution A1

This question in 0606/21 May/June 2022

11 A 5-digit code is to be formed using 5 different numbers selected from 1, 2, 3, 4, 5, 6, 7, 8. Find how many possible codes there are if the code forms (a) a number less than 60 000 that ends in a multiple of 3, [3] (b) an even number less than 60 000. [3]

6 marks

Mark scheme: 11(a) 1080 3 M2 for a fully correct method e.g. [starts with 1, 2, 4, 5 and ends in 3, 6] 4  6  5  4  2 or 960 and [starts with 3 and ends in 6] 1  6  5  4  1 or 120 OR [ends with 6 and starts with 1, 2, 3, 4, 5] 5  6  5  4  1 or 600 and [ends with 3 and starts with 1, 2, 4, 5] 4  6  5  4  1 or 480 or M1 for a partially correct method equivalent to one of the above two steps 11(b) 2160 3 M2 for a fully correct method e.g. [starts with 1, 3, 5 and ends in 2, 4, 6, 8] 3  6  5  4  4 or 1440 and [starts with 2 or 4 and ends in 6, 8 or one of 2 or 4] 2  6  5  4  3 or 720 OR [ends with 6, 8 and starts with 1, 2, 3, 4, 5] 5  6  5  4  2 or 1200 and [ends with 2, 4 and starts with 1, 3, 5 or one of 2 or 4] 4  6  5  4  2 or 960 or M1 for a partially correct method equivalent to one of the above two steps

This question in 0606/21 Oct/Nov 2022

6 A 4-digit code is to be formed using 4 different numbers selected from 2, 3, 4, 5, 6, 7, 8 and 9. Find how many possible codes there are if the code forms (a) a number that is odd and greater than 5000, [3] (b) a number greater than 5000 with a last digit that is prime. [3]

6 marks

Mark scheme: 6(a) 510 3 M2 for a fully correct method e.g. [starts with 5, 7, 9 and ends in 3 or two of 5, 7, 9] 3  6  5  3 or 270 and [starts with 6, 8 and ends in 3,5,7, 9] 2  6  5  4 or 240 OR [ends with 3 and starts with 5, 6, 7, 8, 9] 5  6  5  1 or 150 and [ends with 5, 7, 9 and starts with 6, 8 or two of 5, 7, 9] 4  6  5  3 or 360 or M1 for a partially correct method equivalent to one of the above two steps 6(b) 540 3 M2 for a fully correct method e.g. [starts with 5, 7 and ends with 2, 3 and 5 or 7] 2  6  5  3 = 180 and [starts with 6, 8, 9 and ends with 2, 3, 5, 7] 3  6  5  4 = 360 OR [ends with 2, 3 and starts with 5, 6, 7, 8, 9] 5  6  5  2 or 300 and [ends with 5, 7 and starts with 6, 8, 9 and 5 or 7] 2  6  5  4 or 240 or M1 for a partially correct method equivalent to one of the above two steps

This question in 0606/22 Oct/Nov 2022

5 (a) (i) A gardening group has 20 members. A committee of 6 members is to be selected. Anwar and Bo belong to the gardening group and at most one of them can be on the committee. How many different committees are possible? [2] (ii) The gate for the garden has a lock with a 6-character passcode. The passcode is to be made from Letters G A R D E N Numbers 0 1 2 3 4 5 6 7 8 9. No character may be used more than once in any passcode. Find the number of possible passcodes that have 4 letters followed by 2 numbers. [2] (b) (i) Given that n H 4 , show that ( n - 3) # n C 3 n = 4 # C 4 . [2] = 5n , where n H 3 , show that n satisfies the equation n - 3n - 28 = 0 . (ii) Given that nC 3 2 Hence find the value of n. [4]

10 marks

Mark scheme: 5(a)(i) 35700 2 M1 for 20C6  18C 4 or 18C6  18C5  2C1 oe 5(a)(ii) 32400 2 M1 for 6 P4  10 P2 or ( 6  5  4  3)  (10  9) oe 5(b)(i) Correct algebraic method to show 2 n n ! n B1 for C3  or ( n  3) C3 is the same as 4nC 4 oe 3!( n  3)! n n ! C 4  4!( n  4)! 5(b)(ii) n ( n  1)( n  2) B2 n n ( n  1)( n  2)  5n or B1 for   C3   or 6 6 n(n – 1)(n – 2) = 30n n(n – 1)(n – 2) = 30n seen and completion to given answer: n 2  3 n  28  0 ( n  7)( n  4)  0 oe M1 n = 7 A1

This question in 0606/23 May/June 2023

5 There are 3 women, 2 men and 4 children in a choir. (a) The choir stands in a single straight line. (i) Find the number of possible arrangements if the first person and last person are both women. [2] (ii) Find the number of possible arrangements if all the children stand next to each other. [2] (b) Four of the choir are selected to sing in a group. (i) Find the number of different selections if no man is chosen. [2] (ii) Find the number of different selections if at least 2 women are chosen. [2]

8 marks

Mark scheme: 5(a)(i) 30 240 2 M1for 3  7!  2 or 3 P1  7 P7  2 P1 oe 5(a)(ii) 17 280 2 M1for 4!  6! oe or 4 P4  6 P5 oe 5(b)(i) 35 2 M1 for 7C4 or 1 + 4 + 18 + 12 5(b)(ii) 51 2 M1 for 3C2  6C2 + 3C3  6C1 oe or 18 + 4 + 3 + 2 + 24

This question in 0606/21 May/June 2024