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 :

Total ways = $$3 \times 3 \times 3 \times 3 = 3^4 = 81$$ ways

We choose 1 bag to remain empty ($$\binom{3}{1}$$ ways). We then distribute our 4 books into the remaining 2 bags ($$2^4$$ ways).

$$\binom{3}{1} \times 2^4 = 3 \times 16 = 48$$ ways.

Because of how combinations overlap, when we subtracted the 1 bag empty scenarios, we subtracted the 2 bags empty scenarios twice. We need to add them back once to balance the equation.

$$=\binom{3}{2} \times 1^4 = 3 \times 1 = 3$$ ways.

$$N = 81 - 48 + 3$$

$$N = 36$$