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I have 128 chairs, and 256 people. How many different combinations of the 256 people can be sitting in the 128 chairs? Order doesn't matter, and obviously the same person can't be sitting in more than one chair at a time (no duplicate values in combinations).

Just to clarify, for example, if person 1 is in chair 1 and person 2 is in chair 2 in a given combination, then person 2 being in chair 1 and person 1 being in chair 2 should only count as 1 combination, since they have already been in the same grouping once. I hope that makes sense!

Thanks!

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  • $\begingroup$ I do not know precisely what you are after. There are $\binom{256}{128}$ ways to choose who gets the privilege to sit, if all chairs must be occupied. $\endgroup$ – André Nicolas Jul 7 '15 at 20:30
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It appears as if you are asking how many ways are there to choose 128 people out of the original 256? The solution would simply be ${256 \choose 128}$

For more on combinations, see: http://www.unco.edu/NHS/mathsci/facstaff/Roberson/CourseDocs/MATH%20182/Activities/Combinations%20and%20Permutations.pdf

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  • $\begingroup$ Thank you! Could you possibly extrapolate that calculation for me - I'm a math noob and am not quite sure what that notation means. $\endgroup$ – nicktendo Jul 7 '15 at 20:33
  • $\begingroup$ Of course: The definition ${n \choose x}$ is defined as $\frac{n!}{(n-x)!(x!)}$ Therefore, for this case, the value would be: $\frac{256!}{(128!)(128!)} = 5768658823449206338089748357862286887740211701975162032608436567264518750790$ $\endgroup$ – pMarkov Jul 7 '15 at 20:36

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