Distributing Four Books into Three Bags

To find the number of ways to distribute $4$ distinct books into $3$ distinct bags such that no bag is empty, we can use the Principle of Inclusion-Exclusion or group distribution.

Using the Principle of Inclusion-Exclusion, the total number of ways to distribute the books into the bags without restrictions is $3^4$. Subtracting the cases where at least one bag remains empty gives:

$\text{Number of ways} = 3^4 - \binom{3}{1} 2^4 + \binom{3}{2} 1^4$

Calculating the value:

$\text{Number of ways} = 81 - 3 \times 16 + 3 \times 1 = 81 - 48 + 3 = 36$

Alternatively, partition $4$ distinct books into $3$ non-empty groups, which must have sizes $2, 1, 1$. The number of ways to form these groups and distribute them to $3$ distinct bags is given by:

$\left( \frac{4!}{2! 1! 1! 2!} \right) \times 3! = 6 \times 6 = 36$

Thus, the correct option is B ($36$).