6 (a) Five of the digits 1 2 3 4 5 6 7 8 9 are used to make a 5-digit number. Find how many ways this can be done when the 5-digit number is greater than 50 000. [2] (b) In a group of 13 people, 7 have red hair and 6 have brown hair. (i) The 13 people stand in line for a photograph. No person with red hair is standing next to another person with red hair. Find the number of different ways this can be done. [2] (ii) Chris chooses 5 people from this group. Find the number of ways Chris can do this if at least 1 person chosen has brown hair. [2] Ĭ Ċ Ġ í õ Ī Ċ ü þ

6(a) 8400 2 M1 for 5  8  7  6  5 oe 6(b)(i) 3 628 800 2 M1 for 7!  6! 6(b)(ii) 1266 2 M1 for 13 5 C – 7 5 C

6 (a) (i) Find how many different 5-digit numbers can be formed using the digits 1, 3, 5, 6, 8 and 9. No digit may be used more than once in any 5-digit number. [1] (ii) How many of these 5-digit numbers are odd? [1] (iii) How many of these 5-digit numbers are odd and greater than 60 000? [3] (b) Given that ( ) C C n 45 1 n n 4 1 5 # # = + + , find the value of n. [4]

6 = − c A1 5(a) f ⩾ –4 2 M1for a valid method to find the least value of 2 4 + x x A1 for f ⩾ –4, y ⩾ –4 or f(x) ⩾ –4 5(b) g 1 > B1 Allow 1 > y or ( ) g 1 > x 5(c) ( ) ( )[ ] 2 2 2 1 e 4 1 e 21 + + + = x x M1 4 2 e 6e 16 0 + − = x x ( )( ) 2 2 e 8 e 2 0 + − = x x M1 Dep for quadratic in terms of 2e x and attempt to solve to obtain 2e = x k 2e 2 = x 1 ln 2 = x k M1 Dep on both previous M marks, for attempt to solve 2e = x k ln 2 = x or 1 2 ln2 A1 6(a)(i) 720 B1 6(a)(ii) 480 B1 6(a)(iii) [Starts with 6 or 8]: 192 B1 [Starts with 9]: 72 B1 Total = 264 B1 Alternative [Ends with 9]:48 (B1) [Ends with 1,3 or 5]:216 (B1) Total = 264 (B1) 6(b) ( ) ( )( ) ( ) ( ) 1 1 ! 45 ! 4 !4! 1 5 !5! + + = − + − n n n n n B1 ( ) 2 1 45 5 + = n leading to 15 1 = + n or 2 2 224 0 + − = n n M2 M1 for 15 M1 for n + 1 OR M1 for 2 2 224 0 + − = n n oe M1 for ( )( ) 14 16 0 − + = n n 14 = n only A1

6 (a) A 5-digit number is made using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. No digit may be used more than once in any 5-digit number. Find how many such 5-digit numbers are odd and greater than 70 000. [3] (b) The number of combinations of n objects taken 3 at a time is 2 times the number of combinations of n objects taken 2 at a time. Find the value of n. [4]

6(a) Either Starts with 8: 1680 B1 1680 must not be part of a product. May be implied by final answer Starts with 7 or 9: 2688 B1 May be implied by final answer Total: 4368 B1 Or Alternative 1 Starts with 7, 8 or 9 and ends in 1, 3 or 5: 3024 (B1) Allow for 1008 three times May be implied by final answer Starts with 8 or 9 and ends in 7: 672 Starts with 7 or 8 and ends in 9: 672 (B1) For both May be implied by final answer Total: 4368 (B1) Or Alternative 2 13 ways of obtaining odd 5-digit numbers which start with 7, 8 or 9 (B1) Needs to be part of a product. May be implied by final answer 8 3P ways of arranging the remaining 3 digits: 336 (B1) Needs to be part of a product. May be implied by final answer Total = 13 336 4368 × = (B1) Or Alternative 3 Last digit is 7 or 9: 1344 B1 May be implied by final answer Last digit is 1, 3 or 5: 3024 B1 May be implied by final answer Total: 4368 B1 Or Alternative 4 ( ) ( ) ( ) 10 9 8 8 5 4 3 3 P P 7 P 5 P 4 − × − × − × ( ) 8 3P 5 − × B2 Must be complete Total: 4368 B1 6(b) ( ) ( ) ! 2 ! 3 !3! 2 !2! = − − n n n n soi B1 ( )2 6 − = n soi B2 Dep B1 on first B for ( )2 − n soi Dep B1 on first B for 6 soi 8 = n B1 Dep on previous B marks

7 (a) A six-character password is to be made from the following eight characters. Digits 1 3 5 8 9 Symbols * $ # No character may be used more than once in a password. Find the number of different passwords that can be chosen if (i) there are no restrictions, [1] (ii) the password starts with a digit and finishes with a digit, [2] (iii) the password starts with three symbols. [2] (b) The number of combinations of 5 objects selected from n objects is six times the number of combinations of 4 objects selected from n 1 - objects. Find the value of n. [3]

7(a)(i) 20160 B1 7(a)(ii) 7200 2 B1 for 6 4 P or ( ) 6 5 4 3 360 × × × = for ‘inner’ characters or 5 2 P or ( ) 4 5 20 × = for ‘outer’ characters Must be part of a product 7(a)(iii) 360 2 B1 for 3 3P or 3! or 6 for arrangements of symbols or 5 3P or 5 4 3 × × (= 60) for the digits Must be part of a product 7(b) ( ) ( ) ( ) ( ) 6 1 ! ! 5 !5! 1 4 !4! − = − − − n n n n B1 May be implied by simplification e.g. ( ) 1 ! ! 6 5! 4! − = n n or ( )( )( )( ) 1 2 3 4 5! − − − − n n n n n ( )( )( )( ) 6 1 2 3 4 4! − − − − = n n n n Simplification of either the numerical factorials or the algebraic factorials M1 30 = n A1

7 (a) A committee of 8 people is to be formed from 5 teachers, 4 doctors and 3 police officers. Find the number of different committees that could be chosen if (i) all 4 doctors are on the committee, [2] (ii) there are at least 2 teachers on the committee. [3] (b) Given that P 6 P n n 5 1 4 # = - , find the value of n. [3]

7(a)(i) 8 4 C M1 For realisation that there are 4 places left and 8 people available to fill them 70 A1 7(a)(ii) 1 teacher on committee: 5 ways B1 12 8 C 5 − M1 490 A1 Alternative 2 teachers: 70 3 teachers: 210 4 teachers: 175 5 teachers: 35 (2) B1 for 2 correct cases 490 (B1) 7(b) ( ) ( ) ( ) 1 ! ! 6 5 ! 1 4 ! − = − −− n n n n B1 ( ) ( ) 6 5 ! 5 ! = − − n n n M1 For simplification of either ! n and ( )1 ! − n or ‘cancelling out’ of the terms of ( ) 5 ! − n 6 = n A1 nfww

Cambridge IGCSE Additional Mathematics · 0606

IGCSE Additional Mathematics (0606) Permutations and combinations past paper questions

Permutation and combination questions test counting with and without restrictions, where reading the condition correctly is most of the mark.

46 free questions from 2016–2025, each with its marking scheme. Open the full topical bank to filter by subtopic, paper, and year.

Permutations and combinations in IGCSE Additional Mathematics 0606 exams, 2020–2025

Across 2020–2025, Permutations and combinations accounted for 50 questions worth 337 marks in Cambridge IGCSE Additional Mathematics (0606) papers.