Suppose there are 10 identical red balls and 10 different blue balls. I am wondering if the following is correct. Suppose we want to put those balls into 4 boxes such that:
(a) there is a box containing exactly 2 red balls.
I assume this means that the box contains 2 red balls only and no other balls. So this means, (4 choose 1) to choose the special box. Then the rest of the red balls gets distributed as follows: (10 choose 8). Then the blue balls gets distributed like this: 3^10. So the total is:
4 x (10 choose 8) x (3^10)
(b) there is a box containing exactly 2 blue balls.
to distribute the red, we have 3 boxes. So we have (10 + 3 - 1 choose 10), or (12 choose 10). there are 10 x 9/2! ways of putting balls in the special box that gets 2 blue balls, and 4 ways of choosing the box. Then the rest of the blue balls can be distributed in 3^8 ways. So the total is:
(12 choose 10) x (10 choose 2) x (4) x (3^8) ways.