A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is :
- 18
- 36
- 39
- 72
Answer
(2) 36
Solution
Each of the 4 different books can go into any of the 3 bags: 34=81 ways
Ways with a particular bag empty: 24; with two particular bags empty: 14
By inclusion–exclusion, ways with no bag empty =34-3C1 24+3C2 14=81-48+3=36
Check: books must be split as 2, 1, 1: 4C2 3!=6 6=36
Answer: (2)