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 2 e x and attempt to solve to obtain 2 e = x k 2 e 2 = x 1 ln 2 = x k M1 Dep on both previous M marks, for attempt to solve 2 e = 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

8 (a) A 5-digit number is made using the digits 0, 1, 4, 5, 6, 7 and 9. No digit may be used more than once in any 5-digit number. Find how many such 5-digit numbers are even and greater than 50 000. [3] (b) The number of combinations of n objects taken 4 at a time is equal to 6 times the number of combinations of n objects taken 2 at a time. Calculate the value of n. [5]

8(a) Either Starting with a 6: 120 ways B1 May be implied by final answer Starting with 5, 7 or 9: 540 ways B1 May be implied by final answer Total 660 B1 Or Alternative 1 Ending with a 6: 180 ways (B1) May be implied by final answer Ending with 0 or 4: 480ways (B1) May be implied by final answer Total 660 (B1) Or Alternative 2 11 ways of obtaining even 5-digit numbers which start with 5, 6, 7, 9 (B1) For 11 × k May be implied by final answer 5P3 ways of arranging remaining 3 digits: 60 (B1) For m × 60 where m is from an attempt to list all cases for first and last digits May be implied by final answer 11 60 660 × = (B1) Or Alternative 3 Total arrangements 7P5 minus (all odds + evens starting with 1 + evens starting with 0 or 4) = 2520 – (1440 + 180 + 240) (B2) For 2520 – (1440 + 180 + 240) 660 (B1) 8(b) ( ) ( ) ! 6 ! 4 !4! 2 !2! = − − n n n n B1 ( )( ) 2 3 72 − − = n n 2 B1 for ( )( ) 2 3 − − n n B1 for 72 11 = n only 2 M1 for correct attempt to form and solve a quadratic equation A1 for 11 = n only

4 (a) (i) Find how many different 5-digit numbers can be formed using the digits 1, 2, 3, 5, 7 and 8, if each digit may be used only once in any number. [1] (ii) How many of the numbers found in part (i) are not divisible by 5? [1] (iii) How many of the numbers found in part (i) are even and greater than 30 000? [4] (b) The number of combinations of n items taken 3 at a time is 6 times the number of combinations of n items taken 2 at a time. Find the value of the constant n. [4]

4(a)(i) 720 B1 4(a)(ii) 600 B1 FT on their (i) 5 6 × 4(a)(iii) Starting with 8: 1 × 4 × 3 × 2 × 1 = 24 B1 Starting with 3, 5 or 7: 3 × 4 × 3 × 2 × 2 = 144 M1 May be considering each case separately, need all three cases for M1 A1 Total = 168 A1 4(a)(iii) Alternative Plan for adding numbers ending in 2 and numbers ending in 8 M1 Ending in 2: 1 4 720 96 6 5   × × =     B1 Allow unsimplified Ending in 8: 1 3 720 72 6 5   × × =     B1 Allow unsimplified Total = 168 A1 4(b) 3 2 6 n n C C = B1 ( )( ) 1 2 3! n n n − − ( )( ) ( ) 1 2 6 1 3! 2! n n n n n − − − = B1 ( ) 6 1 2! n n − ( ) ( ) 1 2 18 0 n n n −  − − =   M1 Valid attempt to solve, must have at least one previous B mark 20 n = A1 4(b) Alternative 3 2 6 n n C C = ( ) ( ) 2 !2! 3 !3! n n − = − B1 For dealing with ( ) 2 ! n − and ( ) 3 ! n − to obtain ( ) 2 n − ( ) 2 6 3 n − = × B1 For dealing with 2! and 3! To obtain 6 20 n = M1 Valid attempt to solve, must have at least one previous B mark A1

Cambridge O Level Additional Mathematics · 4037

O-Level Additional Mathematics (4037) Permutations and combinations past paper questions

Revise O-Level Additional Mathematics (4037) — Permutations and combinations with real Cambridge past-paper questions grouped by topic, each shown with its marking scheme. Working one topic at a time targets exactly the skill each question tests.

24 free questions from 2016–2021, each with its marking scheme. Open the full topical bank to filter by subtopic, paper, and year.

Permutations and combinations in O-Level Additional Mathematics 4037 exams, 2020–2025

Across 2020–2025, Permutations and combinations accounted for 28 questions worth 191 marks in Cambridge O Level Additional Mathematics (4037) papers.