The problem states that a teacher wants to divide 15 students into 3 equal groups. The question asks how many ways can the teacher divide the students into 3 groups such that each group has the same number of students.

1. Problem Description

The problem states that a teacher wants to divide 15 students into 3 equal groups. The question asks how many ways can the teacher divide the students into 3 groups such that each group has the same number of students.

2. Solution Steps

First, determine the size of each group. Since there are 15 students and 3 groups, each group must have $15/3 = 5$ students.

Next, we will use combinations to determine the number of ways to form each group.

The number of ways to choose the first group of 5 students from 15 is $\binom{15}{5}$.

After forming the first group, there are $15-5 = 10$ students remaining.

The number of ways to choose the second group of 5 students from the remaining 10 is $\binom{10}{5}$.

After forming the second group, there are $10-5 = 5$ students remaining.

The number of ways to choose the third group of 5 students from the remaining 5 is $\binom{5}{5} = 1$.

Therefore, the total number of ways to form the three groups is $\binom{15}{5} \times \binom{10}{5} \times \binom{5}{5}$. Since the order of the groups does not matter, we divide by $3!$ to avoid overcounting.

So the total number of ways is $\frac{\binom{15}{5} \times \binom{10}{5} \times \binom{5}{5}}{3!}$.

$\binom{15}{5} = \frac{15!}{5!10!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$

$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

$\binom{5}{5} = \frac{5!}{5!0!} = 1$

$3! = 3 \times 2 \times 1 = 6$

$\frac{3003 \times 252 \times 1}{6} = \frac{756756}{6} = 126126$

3. Final Answer

126126