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8.7.6/3! = 56

Here is why this works. There are 8 choices for the fist member of the committee.

Then we have 7 choices for the next and 6 for the last.

HOWEVER, when picking a committee, the order does not matter.

So for example if we pick Chuck, Judy and Melanie and we denote them as C, J and M respectively, then one committee is CJM but that is the same committee as CMJ and MCJ. In fact, there are 3! or 3x2x1=6 ways to arrange the 3 members so we must divide 8x7x6 by 6 and the result is 56 committees of 3 people picked from 8.

Picking a committee is an example of a combination problem in math. If order mattered it would be a permutation.

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Q: How many different 3 person committees can be formed from a group of 8?
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