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. **Choose Books for S1**:
- We need to select 2 books from the 10 available books for student S1. The number of ways to do this is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
- Thus, the number of ways to choose 2 books for S1 is:
\[ \binom{10}{2} \]

2. **Choose Books for S2**:
- After S1 has received 2 books, there are now 8 books left. We need to choose 2 books for student S2 from these remaining 8 books.
- The number of ways to choose 2 books for S2 is:
\[ \binom{8}{2} \]

3. **Choose Books for S3**:
- After distributing books to S1 and S2, there are now 6 books left. We need to choose 3 books for student S3 from these 6 books.
- The number of ways to choose 3 books for S3 is:
\[ \binom{6}{3} \]

4. **Assign Remaining Books to S4**:
- Finally, S4 will receive the remaining books. Since S4 is to receive 3 books and there are exactly 3 books left, there is only one way to assign these books to S4:
\[ \binom{3}{3} = 1 \]

5. **Calculate Total Ways**:
- Now, we multiply the number of ways to distribute the books to each student:
\[ \text{Total Ways} = \binom{10}{2} \times \binom{8}{2} \times \binom{6}{3} \times \binom{3}{3} \]

6. **Substitute Values**:
- Calculate each combination:
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
\[ \binom{3}{3} = 1 \]

7. **Final Calculation**:
- Now, we can calculate the total number of ways:
\[ \text{Total Ways} = 45 \times 28 \times 20 \times 1 \]
\[ = 45 \times 28 \times 20 = 25200 \]

### Final Answer:
The total number of ways to distribute the 10 different books among the 4 students is **25200**.