Problem 165

Analysis

This problem involves the combined application of combinations and permutations. We can first determine that the distribution can be divided into $2$, $2$, and $1$ books.

Step-by-step solution

1. First, we will find the number of ways to choose $2$ books out of $5$ for the first two students who will receive $2$ books each. This can be calculated using the combination formula: $C_5^2=\dfrac{5!}{2!(5-2)!}=10$.

2. Next, we have $3$ books remaining (as the first two students have already received their books), and we will choose $2$ books for the second student who will receive $2$ books. Again, using the combination formula: $C_3^2=\dfrac{3!}{2!(3-2)!}=3$.

3. Now, we have considered the distribution of books to the first two students receiving $2$ books each. However, since there are two students receiving the same number of books, we need to divide the result by the number of arrangements of these students ($A_2^2 = 2$).

4. Finally, there is only $1$ book left, and it will be given to the last student. There are $A_3^3 = 3$ ways to choose this last student.

5. Combining these results, we find the total number of distribution methods: $$\frac{C_5^2 C_3^2}{A_2^2} \cdot A_3^3 = \frac{10 \cdot 3}{2} \cdot 3 = 90$$

Therefore, there are $\boxed{90}$ different methods to distribute the books.