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Q: What is 5 over 4 as a mixed number?

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This is a combination problem because you have to assess a subgroup without a particular order. I find it easiest to first deduce what the total number of arrangements are and then subtract from there. Firstly, There are a total of 9 balls getting assigned to a group of 4. So, to get our total, we use 9C4, which equals 9!/(4!*5!)=126. This means there are a total of 126 arrangements regardless of order. However, this number also includes the combination of 4 like-colored balls we're told to avoid. To calculate the number of arrangements for 4 like-colored balls we set up a separate subgroup for each color. For red balls the possible arrangement are 5C4, which is 5!/(4!*1!)=5. For blue balls the possible arrangements are 4C4, which is 4!/(4!*0!)=1 (remember that 0!=1). So the total number of like-colored arrangements is 5+1=6. Now we just subtract the exceptions from the total to get the answer...126-6=120

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