How many ways can 15 students be divided into 3 groups?

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University of North Bengal

• There are $\,\binom{15}{5}\,$ ways to choose five students for group ONE,
• Now, there are $\binom{10}{5}$ ways to choose five students for group TWO,
• That leaves us with only five students: i.e., hence there is only $\,\binom 55 = 1\,$ way to choose the students for group THREE.

Multiplying these factors: $$\binom{15}{5} \times \binom{10}{5} \times \binom 55= \frac{15!}{10!\,5!}\times \frac{10!}{5!\,5!}\times 1 = \frac{15!}{5!\,5!\,5!}\tag{1}$$

Now we need to divide that total by $\,3!,\,$ since the labeling/ordering of the groups (group ONE, group TWO, group THREE) doesn't matter: since there are $3!$ ways of labeling the groups, we need divide the total given in $(1)$ by $\,3!$.

Hence, our final answer is:

$$\text{There are}\;\left(\frac{15!}{5!\,5!\,5!\,3!}\right)\;\text{ways to divide 15 students into three groups of equal size.}$$