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?
$$\frac{n!}{\left \lfloor {n/2} \right \rfloor!(n – \left \lfloor {n/2} \right \rfloor)!} = \frac{n!}{\left \lceil {n/2} \right \rceil!(n – \left \lceil {n/2} \right \rceil)!}$$
Best Answer
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.