Question 486053
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That would depend on whether the generally accepted assumption that any person who is not female is male (or vice versa) is actually true.  Accepting this initial premise leads us to the conclusion that there are 13 males in the class, and we want to calculte 13 pick 6.


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



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(13\cr\ 6\right\)\ =\ \frac{13!}{6!(13\,-\,6)!}\ =\ \frac{13\times 12\times 11\times 10\times 9\times 8}{6\times 5\times 4\times 3\times 2}\ =\ 13\ \times\ 2\ \times\ 11\ \times\ 2\ \times\ 3]


You can do the rest of the arithmetic yourself.


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