In how many ways can 18 differet books be divided equally among three students ?

Answer

To solve the problem of how many ways 18 different books can be divided equally among three students, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Distribution**: We have 18 different books and we want to divide them equally among 3 students. Each student will receive 6 books.

2. **Selecting Books for the First Student**: We start by selecting 6 books for the first student (let's call them S1) from the 18 available books. The number of ways to choose 6 books from 18 is given by the combination formula:
\[ \text{Ways for S1} = \binom{18}{6} \]

3. **Selecting Books for the Second Student**: After selecting 6 books for S1, we have 12 books left. Now, we need to select 6 books for the second student (S2) from these remaining 12 books:
\[ \text{Ways for S2} = \binom{12}{6} \]

4. **Books for the Third Student**: After distributing books to S1 and S2, there are 6 books left, which will automatically go to the third student (S3). There is only 1 way to give these remaining books to S3:
\[ \text{Ways for S3} = \binom{6}{6} = 1 \]

5. **Calculating Total Ways**: The total number of ways to distribute the books is the product of the ways to choose books for each student:
\[ \text{Total Ways} = \binom{18}{6} \times \binom{12}{6} \times \binom{6}{6} \]

6. **Adjusting for Indistinguishable Groups**: Since the students are indistinguishable in terms of the groups they form (the order in which we select the students does not matter), we need to divide by the number of ways to arrange the 3 groups (which is \(3!\)):
\[ \text{Final Count} = \frac{\binom{18}{6} \times \binom{12}{6} \times \binom{6}{6}}{3!} \]

7. **Substituting Values**: Now we can substitute the values of the combinations:
\[ \binom{18}{6} = \frac{18!}{6! \cdot 12!}, \quad \binom{12}{6} = \frac{12!}{6! \cdot 6!}, \quad \binom{6}{6} = 1 \]
Thus, the final expression becomes:
\[ \text{Final Count} = \frac{\frac{18!}{6! \cdot 12!} \times \frac{12!}{6! \cdot 6!} \times 1}{3!} \]

8. **Simplifying the Expression**: The \(12!\) cancels out, and we are left with:
\[ \text{Final Count} = \frac{18!}{(6!)^3 \cdot 3!} \]

### Final Answer:
The number of ways to divide 18 different books equally among three students is given by:
\[ \frac{18!}{(6!)^3 \cdot 3!} \]