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 $
