AY2021 June Revision P&C Solutions

1. The two empty seats have to be considered as two identical objects, hence \frac{9!}{2!} = 181440 ways.

2. \frac{\binom{11}{3} \binom{8}{3} \binom{5}{3} \binom{2}{2} }{3!} = 15400

3. 4! \times ^5\text{P}_2 = 480 or 6! - 5! \times 2! = 480

4. 26^2 \times 9 \times 8 = 48672

5. (a) 4! \times 5! = 2880

5. (b) 4! \times 3 \times 4! = 1728; \frac{(10-1)!}{3!} = 60480

6. (i) 8! \times ^9 \text{C}_2 \times 2! \times 2! = 5806080 or 10! \times 2 - 9! \times 2! \times 2 = 5806080

6. (ii) (5-1)! \times 3! \times 4! \times 3! \times 12 = 248832

6. (iii) Case 1: 4 distinct letters. ^4 \text{C}_4 \times 4! = 24

Case 2: Exactly 3 distinct letters. ^2 \text{C}_2 \times ^3 \text{C}_2 \times \frac{4!}{2!} =72

Case 3: Exactly 2 distinct letters. ^4 \text{C}_1 \times 3 + ^4 \text{C}_2 = 18

Total number of ways = 24 + 72 + 18 = 114

7. (a) He has two options per coin, thus 2^6 - 1 = 63

7. (b) He has four options, two options and 3 options for each denomination respectively, thus 4 \times 2 \times 3 - 1 = 23

8. (i) If two particular canisters must not be placed on any of the 4 bases, then 6 \times 5 \times 8! = 1209600

8. (ii) If two particular canisters must not be placed next to each other on the same side, then 3628800 - 2 \times 4 \times 2! \times 8! = 2983680

9. (i) \frac{8!}{3!} = 6720

9. (ii) 6720 - \frac{7!}{3!} \times 2! = 5040

9. (iii) Observe there are only two possible cases: D_G_T_S_ or _D_G_T_S.

Total number of ways = 4! \times 4 \times 2! = 192

9. (iv) Observe there are four possible cases: I _ _ I _ _ _ I // I _ _ _ I _ _ I // _ I _ _ I _ _ I // I _ _ I _ _ I _

Total number of ways = 5! \times 4 = 480

10. (i) Case 1: First digit is ‘1’ or ‘2’. \frac{5!}{2} \times 2 = 120

Case 2: First digit is ‘3’: \frac{5!}{2!2!} = 30

Total number of ways = 120 + 30 = 150

10. (ii) Case 1: First digit is ‘1’ and last digit is ‘1’: \frac{4!}{2!} = 12

Case 2: First digit is ‘1’ and last digit is ‘3’: \frac{4!}{2!} = 12

Case 3: First digit is ‘2’ and last digit is ‘3’: \frac{4!}{2!} = 12

Case 4: First digit is ‘2’ and last digit is ‘1’: 4! = 24

Total number of ways = 12 + 12 + 12 + 24 = 60

11. (a) n! \ge 25 \Rightarrow n \ge 5

11. (bi) Method 1: Case 1: all distinct: ^4 \text{C}_3 \times 3! = 24

Case 2: all same: ^3 \text{C}_1 = 3

Case 3: AAB in any order: ^3 \text{C}_1 \times ^4 \text{C}_2 \times \frac{3!}{2!} = 27

Total = 24 + 3 + 27 = 54

Method 2: Case 1: Nobody get hazelnut: 3 \times 3 \times 3 = 27

Case 2: Somebody gets hazelnut: ^3 \text{C}_1 \times 3 \times 3 = 27

Total = 27 + 27 = 54

11. (bii) Method 1: Complement

Number of ways a particular friend gets the hazelnut bar = ^3 \text{C}_2 \times 2! + ^3 \text{C}_1 = 9

Required number of ways = 54 - 9 = 45

Method 2: Direct

Case 1: Nobody gets hazelnut: 3 \times 3 \times 3 = 27

Case 2: Somebody gets hazelnut: ^2 \text{C}_1 \times 3 \times 3 = 18

Total = 27 + 18 = 45

12. (i) Case 1: First digit is ‘3’, last digit can be ‘1’, ‘3’, or ‘5’: 1 \times 3 \times 2 \times 2 = 12

Case 2: First digit is ‘4’, last digit can be ‘1’, ‘3’, or ‘5’: 1 \times 3 \times 2 \times 3 = 18

Case 3: First digit is ‘5’, last digit can be ‘1’ or ‘3’: 1 \times 3 \times 2 \times 2 = 12

Number of ways = 12 + 18 + 12 = 42

12. (ii) 3 \times 5 \times 5 \times 3 - 1 = 224

13. (i) 7! = 5040

13. (ii) ^{10} \text{C}_3 \times \frac{7!}{3!2!} = 50400

13. (iii) \frac{10!}{3!2!} - ^8 \text{C}_3 \times \frac{7!}{2!} = 161280 <!-- /wp:paragraph -->  <!-- wp:paragraph --> 14. (i)(7-1)!7! = 3628800 <!-- /wp:paragraph -->  <!-- wp:paragraph --> (ii)\frac{^6 C_3 \times [(3-1)!]^2 \times 6!}{2} = 28800 $

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