Question:

Step 1: Understand the problem of distribution. The problem describes the distribution of 10 distinct books among four students, let's call them $S_1, S_2, S_3,$ and $S_4$. Each student receives a specified number of books. Step 2: Identify the number of books assigned to each student. According to the problem statement: Student $S_1$ receives 2 books. Student $S_2$ receives 2 books. Student $S_3$ receives 3 books. Student $S_4$ receives 3 books. The total number of books distributed is $2 + 2 + 3 + 3 = 10$, which matches the total number of distinct books available. Step 3: Apply the formula for distributing distinct items into distinct groups. When $n$ distinct items are to be distributed into $k$ distinct groups such that the first group receives $n_1$ items, the second group receives $n_2$ items, ..., and the $k$-th group receives $n_k$ items (where $\sum_{i=1}^{k} n_i = n$), the number of ways to do this is given by the multinomial coefficient formula: $$ \frac{n!}{n_1! n_2! \dots n_k!} $$ Step 4: Calculate the total number of ways to distribute the books. In this case, $n=10$ (total distinct books), and the group sizes are $n_1=2$ (for $S_1$), $n_2=2$ (for $S_2$), $n_3=3$ (for $S_3$), and $n_4=3$ (for $S_4$). Substituting these values into the formula: $$ \text{Number of ways} = \frac{10!}{2! \cdot 2! \cdot 3! \cdot 3!} $$ The final answer is $\frac{10!}{2! \cdot 2! \cdot 3! \cdot 3!}$. Note: The provided question is about complex numbers, but the original solution is for a combinatorics problem. This solution directly addresses the logic provided in the original solution snippet.