Question 399184
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It depends. Is it a committee of 3 equals, such that a committee consisting of Alice, Bill, and Julia is exactly the same as Julia, Alice, and Bill?  Or is the order in which the people are specified significant?

If it is the first way (order doesn't matter) then you need the number of combinations of 8 things taken 3 at a time.

The number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time, *[tex \LARGE \left(n\cr k\right\)], is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]

On the other hand, if order matters, then you need the number of permutations of 8 things taken 3 at a time.  The calculation is very similar except that you remove one factor from the denominator, namely the number of ways you can arrange *[tex \Large k] things.  Hence: *[tex \Large \frac{n!}{(n\,-\,k)!}]

Happy arithmetic.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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