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?

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:

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:

So all together:

To find the number of ways that all the students with red hair end up in the same class, we need to consider the distribution of the remaining students in the other two classes.

Since there are 15 students distributed evenly among 3 classes, each class will have ( \frac{15}{3} = 5 ) students.

If all students with red hair are to end up in the same class, we can treat them as a single entity. Therefore, there are ( \binom{3}{1} = 3 ) ways to choose which class they will be in.

Once the class for the red-haired students is chosen, the remaining students need to be distributed among the remaining two classes. Each of the remaining classes needs to have 5 students, so we need to distribute the remaining 12 students among the 2 classes.

Using the formula for combinations, the number of ways to distribute 12 students into 2 classes of 5 each is ( \binom{12}{5} = 792 ).

Therefore, the total number of ways that all the students with red hair end up in the same class is ( 3 \times 792 = 2376 ).