# 15 students in a school are distributed evenly among 3 classes. Given there are 3 students with red hair in the 15 and distribution is random, how many number of ways that all the students with red hair end up in the same class?

16,632

#### Explanation:

We can force the three red-haired students to be in a class together, leaving room for 2 more students out of the 12 left:

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

$\left(\begin{matrix}12 \\ 2\end{matrix}\right) = 66$ different ways the remaining two students can go into class with the red-haired kids.

Now to deal with the remaining 10 kids. For each of the 66 arrangements in the class with the red-haired kids, we can fill the other two classes with different groups of kids. If we focus on one classroom and choose 5 from the remaining 10, then the last classroom will automatically have the remaining 5 kids. So that's:

$\left(\begin{matrix}10 \\ 5\end{matrix}\right) = 252$

So all together:

$66 \times 252 = \text{16,632}$