# Exercise C.1.7

To choose $k$ objects from $n$, you can make one of the objects distinguished and consider whether the distinguished object is chosen. Use this approach to prove that:

$$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$

Given the distinguished object is chosen, there are $\binom{n-1}{k-1}$ ways to choose the others. If it is not chosen, there are $\binom{n-1}{k}$ ways to choose the objects. Adding those together yields the equality above.

\begin{aligned} \binom{n-1}{k} + \binom{n-1}{k-1} &= \frac{(n-1)!}{k!(n-1-k)!} + \frac{(n-1)!}{(k-1)!(n-1-k+1)!} \\ &= \frac{(n-1-k+1)(n-1)!}{k!(n-1-k+1)!} + \frac{k(n-1)!}{k!(n-1-k+1)!} \\ &= \frac{(n-1-k+1+k)(n-1)!}{k!(n-1-k+1)!} \\ &= \frac{n!}{k!(n-k)!} \\ &= \binom{n}{k} \end{aligned}