# AY2021 June Revision P&C Solutions 1. The two empty seats have to be considered as two identical objects, hence ways.

2. 3. or 4. 5. (a) 5. (b) ; 6. (i) or 6. (ii) 6. (iii) Case 1: 4 distinct letters. Case 2: Exactly 3 distinct letters. Case 3: Exactly 2 distinct letters. Total number of ways 7. (a) He has two options per coin, thus 7. (b) He has four options, two options and 3 options for each denomination respectively, thus 8. (i) If two particular canisters must not be placed on any of the 4 bases, then 8. (ii) If two particular canisters must not be placed next to each other on the same side, then 9. (i) 9. (ii) 9. (iii) Observe there are only two possible cases: D_G_T_S_ or _D_G_T_S.

Total number of ways 9. (iv) Observe there are four possible cases: I _ _ I _ _ _ I // I _ _ _ I _ _ I // _ I _ _ I _ _ I // I _ _ I _ _ I _

Total number of ways 10. (i) Case 1: First digit is ‘1’ or ‘2’. Case 2: First digit is ‘3’: Total number of ways 10. (ii) Case 1: First digit is ‘1’ and last digit is ‘1’: Case 2: First digit is ‘1’ and last digit is ‘3’: Case 3: First digit is ‘2’ and last digit is ‘3’: Case 4: First digit is ‘2’ and last digit is ‘1’: Total number of ways 11. (a) 11. (bi) Method 1: Case 1: all distinct: Case 2: all same: Case 3: AAB in any order: Total Method 2: Case 1: Nobody get hazelnut: Case 2: Somebody gets hazelnut: Total 11. (bii) Method 1: Complement

Number of ways a particular friend gets the hazelnut bar Required number of ways Method 2: Direct

Case 1: Nobody gets hazelnut: Case 2: Somebody gets hazelnut: Total 12. (i) Case 1: First digit is ‘3’, last digit can be ‘1’, ‘3’, or ‘5’: Case 2: First digit is ‘4’, last digit can be ‘1’, ‘3’, or ‘5’: Case 3: First digit is ‘5’, last digit can be ‘1’ or ‘3’: Number of ways 12. (ii) 13. (i) 13. (ii) 13. (iii) (7-1)!7! = 3628800 \frac{^6 C_3 \times [(3-1)!]^2 \times 6!}{2} = 28800 \$

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