# In how many different ways can I sort balls of two different colors

Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I can place them? And what if I also have 3 red balls?

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maybe a title more clear could be "In how many different ways can I sort balls of two different colors" –  mau Jul 22 '10 at 7:37
So, you're asking for how many unique ways you can sort them? And swapping two red balls in a certain arrangement doesn't count as a unique new sorting. –  Justin L. Jul 22 '10 at 7:38
@Justin L., That's probably a fair assumption, yeah. –  Noldorin Jul 22 '10 at 7:53

This is a standard problem involving the combinations of sets, though perhaps not very obvious intuitively.

Firstly consider the number of ways you can rearrange the entire set of balls, counting each ball as indepndent (effectively ignoring colours for now). This is simply (4 + 5)! = 9!, since the 1st ball can be any of the 9, the 2nd can be any of the remaining 8, and so on.

Then we calculate how many different ways the yellow balls can be arranged within themselves, since for the purpose of this problem they are considered equivalent. The number of combinations is of course 4!; similarly, for the blue balls the number is 5!.

Hence, overall we find:

total no. of arrangements = number of arrangements of all balls / (number of arrangements of yellow balls * number of arrangements of blue balls)

Therefore in our case we have:

total no. of arrangements = 9! / (5! * 4!) = 126

I'm sure you can see how this can be easily extended if we also have 3 red balls too. (Hint: the total changes and we have another multiple of identical arrangements to account for.)

Hope that helps...

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