Question 524688
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A bunch.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{k\,=\,1}^{52}\ \left(52\cr\ k\right\)] 


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


You can save yourself half the work by using symmetry:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sum_{k\,=\,1}^{26}\ \left(52\cr\ k\right\)]


Also, in Excel (on the PC) or Numbers (on a Mac) the formula


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =COMBIN(n,k)]


yields *[tex \LARGE \left(n\cr k\right\)]


just fill in the numbers for *[tex \LARGE n] and *[tex \LARGE k] that you need.


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