Exercise C.1.10

Show that for any n0n \ge 0 and 0km0 \le k \le m, the expression (nk)\binom{n}{k} achieves its maximum value when k=n/2k = \lfloor n/2 \rfloor or k=n/2k = \lceil n/2 \rceil.

It's evident in Pascal's triangle, yet anyway. There are kk multipliers in the denominator of:

(nk)=n!k!(nk)! \binom{n}{k} = \frac{n!}{k!(n-k)!}

It's a question of minizing them. For n/2k<i<n/2n/2 - k < i < n/2, there would be ii pairs of the type i(ni)i(n-i). Those pairs are strictly greater than i2i^2. They are minized when ii is n/2n/2 or the nearest integer.

It's not bullet-proof, but it works.