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 canisters are near the doors – only 1 canister near the doors=     4 4 6 2 1 110! 2! 8! 2! 8! 1209600C C C        (ii) No. of ways = 8 110! 2! 8! 2983680C    Explanation: No. of ways without restriction – (Choose the positions for the 2 canisters x Permutate the the positions for the canisters x Arrange the other 8 canisters) Method 2: (If Can 1 is on A, on B, on C, …, J)                     8 8! 7 8! 7 8! 7 8! 8 8! 8 8! 7 8! 7 8! 7 8! 8 8! 2983680                     Door Door A B C D E F G H I J

18-T3 4 [N2008/II/10] A group of diplomats is to be chosen to represent three islands, K, L and M. The group is to consist of 8 diplomats and is chosen from a set of 12 diplomats consisting of 3 from K, 4 from L and 5 from M. Find the number of ways in which the group can be chosen if it includes (i) 2 diplomats from K, 3 from L and 3 from M, (ii) diplomats from L and M only, (iii) at least 4 diplomats from M, (iv) at least 1 diplomat from each island. Concepts Tested 1) Splitting into mutual exclusive cases 2) Complement method (i) No. of ways = 3 4 5 2 3 3 120C C C   (ii) Method 1: No. of ways = 9 8 9C  Method 2: Cases No. of ways =     4 5 4 5 4 4 3 5 9C C C C  Method 3: Complement No restriction – (1 K) – (2 K) – (3 K) =       12 3 9 3 9 3 9 8 1 7 2 6 3 5 9C C C C C C C    (iii) Method 1: Cases Case 1: 4 from M No. of ways = 5 7 4 4 175C C  Case 2: 5 from M No. of ways = 5 7 5 3 35C C   Total no. of ways = 175 + 35 = 210 Method 2: Complement No restriction – (0 M) – (1 M) – (2 M) – (3 M) =         12 5 7 5 7 5 7 8 1 7 2 6 3 50 210C C C C C C C        (iv) No. of ways = No. of ways with no restrictions – No. of ways if no K – No. of ways if no L = 12 9 8 8 8 8 485C C C   *Note: There is no case for No. of ways if no M since the other 2 islands only make up 7 diplomats (i.e., there must always be at least 1 diplomat from M) Common mistake: (iii) Fix 4 M, choose remaining 4 people. 5 8 4 4C C This is an overcount. Example: M1, M2, M3, M4, M5 M1, M2, M3, M4 M5, L1, L2, K1 M1, M2, M3, M5 M4, L1, L2, K1 (iv) Consider cases of (1) 1K, (2) 2K, (3) 3K, (4) 1L, (5) 2L, (6) 3L… [overcount] Fix 1K, 1L, 1M, choose remaining 5 people. 3 4 5 9 1 1 1 5C C C C [overcount] Example: K1, L1, M1 K2, L2, L3, L4 K2, L1, M1 K1, L2, L3, L4

18-T4 5 [N14/II/6] A team in a particular sport consists of 1 goalkeeper, 4 defenders, 2 midfielders and 4 attackers. A certain club has 3 goalkeepers, 8 defenders, 5 midfielders and 6 attackers. (i) How many different teams can be formed by the club? One of the midfielders in the club is the brother of one of the attackers in the club. (ii) How many different teams can be formed which include exactly one of the two brothers? The two brothers leave the club. The club manager decides that one of the remaining midfielders can play as either a midfielder or as a defender. (iii) How many different teams can now be formed by the club? Concepts Tested 1) Multiplication Principle 2) Splitting into mutual exclusive cases (i) Number of teams = 3 8 5 6 1 4 2 4 31500C C C C  (ii) Case 1: Number of teams that has the midfielder brother = 3 8 4 5 1 4 1 4 4200C C C C  Case 2: Number of teams that has the attacker brother = 3 8 4 5 1 4 2 3 12600C C C C  Total number of teams = 4200 + 12600 = 16800 (iii) Method 1 Case 1: Number of teams if one of the midfielders (particular) plays as midfielder = 3 8 3 5 1 4 1 4 3150C C C C  Case 2: Number of teams if one of the midfielders (particular) plays as defender = 3 8 3 5 1 3 2 4 2520C C C C  Case 3: Number of teams if one of the midfielders (particular) does not play = 3 8 3 5 1 4 2 4 3150C C C C  Total = 3150 + 2520 + 3150 = 8820 Method 2 3 8 4 5 3 9 3 5 3 8 3 5 1 4 2 4 1 4 2 4 1 4 2 4 8820C C C C C C C C C C C C   (player as Midfielder) + (player as Defender) – (without the player)

18-T5 6 A committee of 5 people is to be chosen from 6 married couples. Find how many ways this committee can be chosen if (a) all are equally eligible, (b) the two youngest women and at most one of the oldest two men are to be included, (c) the committee must consist of at least one woman and at least one man, (d) no husband and wife can serve in the same committee? Concepts Tested 1) Combination with and w/o restrictions 2) Complement method 3) Splitting into mutual exclusive cases (a) Total no. of ways w/o restrictions = 12 5 792C  (b) Method 1: Cases Case 1: 2 youngest women in and the oldest two men not in No. of ways = 2 8 2 3 56C C  Case 2: 2 youngest women in and one of the oldest two men in No. of ways = 2 2 8 2 1 2 56C C C   Hence, total no. of ways = 112 Method 2: Complement No. of ways