Question 573959
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If you consider all of the possibilities for various combinations of *[tex \LARGE n] things taken *[tex \LARGE k] at a time and listed them in desceding order of *[tex \LARGE k], you would get a list that looks like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,n)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,n\,-\,1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,n\,-\,2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,7)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C\left(n,\,\frac{n}{2}\right)]


(Since *[tex \LARGE C(n,\,7)\ =\ C(n,\,5)], these two list items must be equidistant from the middle item in the list.)



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cdot]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,0)]


But since the *[tex \LARGE k]s must all be integers, there can only be one item in the list between 7 and 5, namely


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C(n,\,6)]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{n}{2}\ =\ 6]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ =\ 12]



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