2024 Permutations and Combinations Tutorial Solutions

18-T1 Suggested Anchor Questions: Q4, 7, 8, 11 Section 1: Discussion Questions (Students are to attempt all questions in this section) 1 A delegation of 7 people is to be selected from a group of 10 men and 8 women. Find the number of such delegations that contain at least one man. 2 (a) In how many ways can five different books be distributed among 10 people if each person can get any number of books? (b) Mary has seven cousins altogether. In how many ways can she invite some or all of them to her birthday party next Saturday? Concepts Tested 1) Multiplication Principle (a) Each book can be distributed in 10 1C ways, so 10 10 10 10 10 1 1 1 1 1total 100000 waysC C C C C      (b) Method 1: Each cousin can either be invited or not invited — i.e 2 ways. So total no. of ways 72 1 128 1 127     Method 2: No. of ways 7 7 7 7 7 7 7 1 2 3 4 5 6 7 127C C C C C C C        Common mistake: (a) Consider the 10 people rather than the 5 books to be distributed. (b) 72 128 [Not removing the case where none is invited.] H2 Mathematics (9758) Topic 18: Permutations & Combinations Tutorial Questions Concepts Tested 1) Combination using Complement method Method 1: Complement No. of delegations that contain at least one man = Total w/o restrictions – Delegations with all women = 18 8 7 7C C 31816  Method 2: Direct               10 8 10 8 10 8 10 8 10 8 10 8 10 8 1 6 2 5 3 4 4 3 5 2 6 1 7 0 31816C C C C C C C C C C C C C C       Common mistake: Fix 1 man, choose remaining 6 people. 10 17 1 6 123760C C  This is an overcount: for example, 1 2 1 2 3 4 5, , , , , ,M M W W W W W and 1 2 3 4 52 1, , , , , ,M M W W W W W can both happen, but they are the same selection of people actually.

18-T2 3 [N2004/II/4] A rectangular shed, with a door at each end, contains ten fixed concrete bases marked A, B, C, …, J, five on each side (see diagram). Ten canisters, each containing a different chemical, are placed with one canister on each base. In how many different ways can the canisters be placed on the bases? Find the number of ways in which the canisters can be placed (i) if 2 particular canisters must not be placed on any of the 4 bases A, E, F and J next to a door, (ii) if 2 particular canisters must not be placed next to each other on the same side of the shed. Concepts Tested (i) Permutation with restrictions (ii) Permutation using Complement Method No. of ways in which canisters can be placed in = 10! 3628800 (i) Method 1: Consider the 2 particular canisters first No. of ways = 6 2 2! 8! 1209600C    Method 2: Put 4 other canisters at A, E, F, J first No. of ways = 8 4 4! 6! 1209600C    From the remaining 8 canisters, pick 4 to put at the 4 corners then fill the 6 bases in the centre with 6 canisters. Method 3: Complement No. of ways without restriction – both caniste