The problem is to show that if $n$ is a positive integer, then $$1 = C(n, 0) < C(n, 1) < ... < C(n, \left \lfloor{ n/2} \right \rfloor) = C(n, \left \lceil {n/2} \right \rceil) > ... > C(n, n-1) > C(n, n) = 1$$.

Where $C(n, k) = \frac{n!}{k!(n-k)!}$

How can I prove the following to be equal, where n is a positive integer?

$$\frac{n!}{\left \lfloor {n/2} \right \rfloor!(n - \left \lfloor {n/2} \right \rfloor)!} = \frac{n!}{\left \lceil {n/2} \right \rceil!(n - \left \lceil {n/2} \right \rceil)!}$$


Consider odd and even $n$ separately. If $n$ is even, write $n=2k$ and compute the floor and ceiling. If $n$ is odd, write $n=2k+1$ and compute the floor and ceiling, then justify the central equality.

  • $\begingroup$ I've ran into a wall in the case where $n$ is odd ($n=2k+1$). I end up with $$\frac{(2k+1)!}{k!(2k-k)!} = \frac{(2k+1)!}{(k+1)!(2k-k+1)!}$$ which doesn't seem to make sense to me if i try to work it out... $\endgroup$ – user11892 Mar 30 '15 at 1:44
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    $\begingroup$ You have two sloppy errors. In the parentheses on the bottom the $2k$ should be $2k+1$ on both sides, and on the right the $-k+1$ should be $-(k+1)$. $\endgroup$ – Ross Millikan Mar 30 '15 at 1:55
  • $\begingroup$ Wow it's been a long day... thank you Mr. Ross $\endgroup$ – user11892 Mar 30 '15 at 1:56

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