# Combinations with repitions when there's limits on how many repitions

1. Sep 7, 2014

### fleazo

Note, this is not a homework problem, as I'm not even in college. I just had a quick question.

I know the formulas to do things such as "How many ways could you choose 5 balls from a tub of yellow, red, and blue colored balls?" (where you envision in this case, a tub where there's more than 5 yellow, more than 5 red, and more than 5 blue)

But what if you have a situation where say, there's more than 5 red and more than 5 blue, but maybe there's only 3 yellow? So there's a "limit" on one of your repetition groups. When you run in to situations like this, do you have to break things down in to disjoint sets?

2. Sep 7, 2014

### Stephen Tashi

Perhaps if you write out an example of such a formula, someone can tell you how to modify it to work in the case when there are only 3 yellow balls.

3. Sep 7, 2014

### fleazo

Sure. So if the task is to determine how many colored 5-ball subsets are possible using a tub of yellow, red, and blue colored balls, this would be (using the formula on this pdf http://www.csee.umbc.edu/~stephens/203/PDF/6-5.pdf):

C(5 + (3-1), 5) = C(7,5) = 7!/(5!(7-5)!) = 7!/(5!2!) = 7*6/2! = 7*3 = 21

Last edited: Sep 7, 2014