Question 448029
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10 ways to pick the first guy in line.  For each of those ways, 9 ways to pick the 2nd guy, hence 10 times 9 = 90 ways to pick the first two guys.  For each of those 90 ways, 8 ways to pick the third guy, so 720 ways to pick the first three guys.  And so on...


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\ \times\ 9\ \times\ \cdots\ \times\ 3\ \times\ 2\ \times\ 1\ =\ 10!]



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