Permutation and combination questions are found in paper 2, section B (Statistics section) of the H2 A Level Math exam. In this post, you will find the worked solutions for past A Level questions on permutation and combination.

A list of past questions from each statistics topic can be found here.

A quick summary of formulae in each statistics topics can be found here.

## Past A Level Permutation and Combination Questions

These are the past A Level permutation and combination questions for the new syllabus (syllabus 9758) which started from 2017.

Click on the link to go straight to the worked solutions for each question.

## Permutation and Combination Question: 2017 Paper 2 Question 6

(i) No. of ways = 5! (4!)^{5} = 95 551 480

(ii) No. of ways = 10!(3!)(3!)(3!)(2) = 1 567 641 600

(iii) Probability = [(15-1)! x 5! x ^{15}C_{5} ]/ (20-1)!= 0.258 (3 s.f.)

Note: part (iii) is a permutation and combination cum probability question.

## Permutation and Combination Question: 2019 Paper 2 Question 6

(i) These 22 clubs are all the clubs in Division One. Since Alice is interested to find out approaches to training by clubs in Division One, and she sent questionnaire to all these 22 clubs, they form a population.

(ii) Dilip should select a sample size he wished to conduct the investigation. The sample should be chosen such that 22% are from Division One, 24% are from Division Two, 26% are from Division Three, and 28% are from Division Four.

For instance, if Dilip wants to study a sample size of 50, he would select randomly 11 clubs from Division One; then randomly select 12 clubs from Division Two, randomly select 13 clubs from Division Three, and randomly select 14 clubs from Division Four.

(iii) No. of ways = ^{22}C_{5} x ^{24}C_{5} x ^{26}C_{5} x ^{28}C_{5} =7.24 x 10^{18} (3 s.f.)

Note: Part (i) and (ii) tests students on their concepts of random samples, population vs sample. Part (iii) is a permutation and combination question.

## Permutation and Combination Question: 2021 Paper 2 Question 7a, b

(a) No. of ways = 11! / (5! 2! 2!) = 83 160

(b) No fo ways = 4! (3)(2!) = 144

(c) Probability = [7!/ (2! 2!)] / 83 160 = 1/ 66

Note: part (iii) is a permutation and combination cum probability question.

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