The number of ways of disturbing 10 different books among 4 students `(S_(1)-S_(4))` such that `S_(1) and S_(2)` gets 2 books and `S_(3) and S_(4)` get 3 books each is:

Options

A. 12600
B. 25250
C. `""^(10)C_(4)`
D. `(10!)/(2!2!3!3!)`

Answer

To solve the problem of distributing 10 different books among 4 students (S1, S2, S3, and S4) such that S1 and S2 each receive 2 books, while S3 and S4 each receive 3 books, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the distribution of books**:
- S1 gets 2 books.
- S2 gets 2 books.
- S3 gets 3 books.
- S4 gets 3 books.

Total books distributed = 2 + 2 + 3 + 3 = 10 books.

2. **Choose books for S1**:
- We need to select 2 books out of the 10 for S1. The number of ways to do this is given by the combination formula:
\[ \text{Ways for S1} = \binom{10}{2} \]

3. **Choose books for S2**:
- After S1 has received 2 books, there are 8 books left. We need to select 2 books for S2 from these remaining 8 books:
\[ \text{Ways for S2} = \binom{8}{2} \]

4. **Choose books for S3**:
- Now, there are 6 books remaining. We need to select 3 books for S3:
\[ \text{Ways for S3} = \binom{6}{3} \]

5. **Assign books to S4**:
- Finally, S4 will receive the remaining 3 books. Since there are exactly 3 books left, there is only one way to give these books to S4:
\[ \text{Ways for S4} = \binom{3}{3} = 1 \]

6. **Calculate the total number of ways**:
- The total number of ways to distribute the books is the product of the ways for each student:
\[ \text{Total Ways} = \binom{10}{2} \times \binom{8}{2} \times \binom{6}{3} \times \binom{3}{3} \]

7. **Substitute the values**:
- Now we can calculate each combination:
\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
\[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]
\[ \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
\[ \binom{3}{3} = 1 \]

8. **Final Calculation**:
- Now, multiply these results together:
\[ \text{Total Ways} = 45 \times 28 \times 20 \times 1 \]
- Calculate:
\[ 45 \times 28 = 1260 \]
\[ 1260 \times 20 = 25200 \]

Thus, the total number of ways to distribute the 10 different books among the 4 students is **25200**.