Given \(n\) distinct items, there are \(P(n, k)\) permutations of \(0 \leq k \leq n\) items. Each permutation of \(k\) items can be re-arranged \(k!\) ways. If item order doesn’t matter, there are
\[ \frac{P(n, k)}{k!} = \frac{n!}{k!(n-k)!} = C(n, k) \]
orderless permutations, or combinations. Sanity check: \(C(n, n) = n!/(n!(n - n)!) = 1/0! = 1\).
Combinations are fun to play with. First, \(C(n, k) = C(n, n - k)\).? Then
\[\begin{eqnarray}
C(n, k + 1) & = & \frac{n!}{(k + 1)!(n - k - 1)!} \\
& = & \frac{n!(n - k)}{(k + 1)k!(n - k)(n - k - 1)!} \\
& = & \frac{n - k}{k + 1}C(n, k)
\end{eqnarray}\]
\(C(n, k + 1)\) is largest when \((n - k)/(k + 1)\) is closest to 1, which occurs when \(k\) is closest to \((n - 1)/2\). Similarly
\[\begin{eqnarray}
C(n + 1, k) & = & \frac{(n + 1)!}{k!(n + 1 - k)!} \\
& = & \frac{(n + 1)n!}{k!(n + 1 - k)(n - k)!} \\
& = & \frac{n + 1}{n + 1 - k}C(n, k)
\end{eqnarray}\]