Question 1163219
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  nPr\ =\ \frac{n!}{(n\,-\,r)!}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  nCr\ =\ \frac{n!}{r!(n\,-\,r)!]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r!\ =\ \frac{nPr}{nCr}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r!\ =\ \frac{336}{56}\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r\ =\ 3]


336 divided by 8 is 42.  Then 6 and 7 are factors of 42.  So the numerator of nPr is 8 X 7 X 6, and then the denominator must be 5! and 8 - 3 is 5, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  n\ =\ 8]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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