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Math I know; Basketball I'm a little weaker on.

Assuming that each of the five positions is distinct there are 8! / 3! or 6720 possible combinations.

Assuming that all of the five positions are the same (I know this isn't true; I'm just putting it up for comparison) there are 8 choose 5 ( 8!/(3!*5!)) or 56 possible combinations.

If some of the positions are the same but others aren't, then the answer is somewhere in between.

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Q: How many ways can the 5 starting positions on a basketball team be filled 8 men who can play any of the positions?

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The first position can be filled by eight, the second by seven, and so on. So the number of possibilities are: 8!/3! = 8*7*6*5*4 = 6720

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Since there were many of them, the positions wer probably filled by "extras."

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