Problem - 4282
Let $n$ and $k$ be two positive integers. Show that $$\frac{1}{\binom{n}{k}}=\frac{k}{k-1}\left(\frac{1}{\binom{n-1}{k-1}}-\frac{1}{\binom{n}{k-1}}\right)$$
Starting from the right side $$\begin{array}{rl} &\displaystyle \frac{k}{k-1}\left(\frac{1}{\binom{n-1}{k-1}}-\frac{1}{\binom{n}{k-1}}\right)\\ \\ =& \displaystyle\frac{k}{k-1}\left(\frac{(k-1)!(n-k)!}{(n-1)!}-\frac{(k-1)!(n-k+1)!}{n!}\right)\\ \\ =&\displaystyle\frac{k}{k-1}\cdot\frac{n(k-1)!(n-k)! - (k-1)!(n-k+1)!}{n!} \\ \\ =& \displaystyle\frac{k}{k-1}\cdot\frac{(k-1)!(n-k)!}{n!}\left(n - (n-k+1)\right) \\ \\ =&\displaystyle\frac{k!(n-k)!}{n!}\\ \\=&\displaystyle\frac{1}{\binom{n}{k}}\end{array}$$