Question 313187
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You need the number of ways to select *[tex \Large k] things from a selection of *[tex \Large n] things where order does not matter.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(n\cr k\right\)\ =\ \frac{n!}{k!(n\,-\,k)!}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(100\cr\ \,5\right\)\ =\ \frac{100!}{5!(100\,-\,5)!}]


Which is some very nasty arithmetic.  However, there is a shortcut...


If you divide *[tex \Large 100!] by *[tex \Large 95!] you get *[tex \Large 100\ \times\ 99\ \times\ 98\ \times\ 97\ \times\ 96]


Then divide this product by *[tex \Large 5!\ =\ 5\ \times\ 4\ \times\ 3\ \times\ 2\ \times\ 1]


The 5 goes into 100 20 times, the 4 goes into 96 24 times, the 3 goes into 99 33 times and the 2 goes into 98 49 times, so the whole thing reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20\ \times\ 33\ \times\ 49\ \times\ 97\ \times\ 24]


Somewhat simpler.  You can take it from here.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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