Question 527586
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That depends.  Are all 15 players equally adept at any of the 10 positions?  The question becomes does order matter, which is to say is Pat in position 1 and Sandy in position 2 a different outcome than Sandy in position 1 and Pat in position 2?


If yes, then the answer is the number of permutations of 15 things taken 10 at a time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15!}{(15\ -\ 10)!}]


If no, then the answer is the number of combinations of 15 things taken 10 at a time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15!}{10!(15\ -\ 10)!}]


You can do your own arithmetic.


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