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I'm just wondering why the n choose k formula

$n \choose k$ = $\frac{n!}{k!(n-k)!}$

always gets you a whole number. I know that obviously there can only be a whole number of "ways" to choose k items in n items, but I'm more interested on how the factorials divide.

staticvoid
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1 Answers1

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Note that $k<n$. So when you write out what $n!$ contains, you will see $k!$ and $(n-k)!$ are included.

Argyll
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    It's reasonably obvious that they're both included separately, but is it really obvious that both are included? – pjs36 Mar 24 '18 at 01:55
  • I'd say it isn't trivial – Dylan Mar 24 '18 at 05:11
  • if a number x is divisible by y and z , that does not mean x/(y*z) is an integer. – cyberboy Mar 24 '18 at 21:15
  • @cyberboy that wasn't the point. – Argyll Mar 25 '18 at 02:40
  • @Argyll , but this is the fault in your reasoning. here x = n! , y = k! and z = (n-k)! , you are saying that x is divided by y , and x is divided by z, but this does not mean x is divisible by y and z simultaneously , which is what we have to prove for showing it as whole number – cyberboy Mar 25 '18 at 10:25
  • Please don't assume what I am saying. – Argyll Mar 25 '18 at 23:41