Simplifying $\binom{n}{k}$ / $\binom{n}{k-1}$

Question

So the question is as follows:

Simplify $$ \frac{\binom{n}{k}}{\binom{n}{k-1}}. $$

And this is what I got: $$\begin{align*} \frac{\binom{n}{k}}{\binom{n}{k-1}} &= \frac{\frac{n!}{(n-k)!k!}}{\frac{n!}{(n-k-1)!(k-1)!}}\\[1em] &= \frac{n!}{(n-k)!k!} \times \frac{(n-k-1)!(k-1)!}{n!}\\[1em] &= \frac{(n-k-1)!(k-1)!}{(n-k)!k!}\\[1em] &= \frac{\left(\frac{(n-k)!}{(n-k)}\right)\left(\frac{k!}{k}\right)}{(n-k)!k!}\\[1em] &= \frac{\frac{(n-k)!k!}{(n-k)k}}{(n-k)!k!}\\[1em] &= \frac{1}{(n-k)k}. \end{align*}$$

However, my textbook says that the answer is

$$\frac{n-k+1}{k}.$$

Can someone please point me in the right direction?

Answer 1

First note that $$ \binom{n}{k}=\frac{n!}{k!(n-k)!}, $$ and $$ \binom{n}{k-1}=\frac{n!}{(k-1)!(n-k+1)!}. $$ Thus, we will have the following: \begin{align} \frac{\binom{n}{k}}{\binom{n}{k-1}} &= \frac{n!}{k!(n-k)!}\cdot\frac{(n-k)!(n-k+1)(k-1)!}{n!}\\[1em] &= \frac{(n-k+1)(k-1)!}{k(k-1)!}\\[1em] &= \frac{n-k+1}{k}. \end{align} This confirms what the book had as the answer.