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or, similarly, given 25 switches, how many ways are there to turn on 5 of them...

I'm not interested in the number, I want to know how to calculate it...

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    $\begingroup$ How many ways are there of choosing 5 elements out of a set of 25? $\endgroup$
    – Gary.
    Commented Jul 19, 2015 at 20:28
  • $\begingroup$ Instinctively I think ... if we have 25 numbers, and we pick at random, it's 25*24*23*22*21. But order doesn't matter (1,2,3,4,5 is the same as 1,2,3,5,4), so it's... less than that? $\endgroup$
    – Squish
    Commented Jul 19, 2015 at 20:38
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    $\begingroup$ 25*24*23*22*21, and then there are 5! identical orderings $\endgroup$
    – qwr
    Commented Jul 19, 2015 at 20:39
  • $\begingroup$ @qwr: Do you want to close the deal and write it down, or should I? $\endgroup$
    – Gary.
    Commented Jul 19, 2015 at 22:26
  • $\begingroup$ @Gary. You can try answering, but the question looks like it'll be closed soon $\endgroup$
    – qwr
    Commented Jul 20, 2015 at 2:27

1 Answer 1

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You are looking for the number of combinations of five things out of twenty-five, written $25 \choose 5$ (there are other ways that are used). You have $25$ squares for the first checker, $24$ for the second, down $21$ for the fifth, but we don't care what order we put the checkers down in, so we divide by $5!$ orders we could have chosen giving ${25 \choose 5}=\frac {25 \cdot 24 \cdot 23 \cdot 22 \cdot 21}{5!}=\frac {25!}{5!20!}$

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