Proving $\binom{n}{r}=\frac{n-r+1}{r}\binom{n}{r-1}$ How would one prove the following using algebra.
$$\binom{n}{r}=\frac{n-r+1}{r}\binom{n}{r-1}$$
This I did but I am stuck.
$$\frac{n-r+1}{r}\frac{n!}{r-1!(n-r-1!)}$$
Then I have
$$\frac{n(n-1)(n-2)(n-3)!(n-r+1)}{(r-1)(r-2)(r-3)!(n-r-1)(n-r-2)!r}$$
But I find myself stuck.
 A: The lefthand side $\binom{n}{r}$ counts the number of committees (where everyone has equal rank) one can form by choosing $r$ people from $n$ available people.
The righthand side counts the same thing. We'll see it in two steps.
First, form a committee of $r-1$ people in $\binom{n}{r-1}$ ways. Next, choose a committee chairperson from the remaining $n-r+1$ people, for a total of $(n-r+1) \binom{n}{r}$ committees with a chair.
But wait. We don't want to count committees with chairs. We want to count committees where everyone is equal. Which committees turn out to be the same if we strip the chair of his/her rank? Within a given committee, there are $r$ ways we could have assigned the role of chair. Therefore, we overcount by a factor of $r$ in the second paragraph. We correct this with division and conclude that there are $\frac{n-r+1}{r}\binom{n}{r-1}$ committees where everyone is equal.
A: $$\begin{align}\frac{n-r+1}{r}\binom{n}{r-1}&=\frac{n-r+1}{r}\cdot\frac{n!}{(r-1)!(n-r\color{red}{+}1)!}\\&=\frac{n!(n-r+1)}{r!(n-r+1)!}\\&=\frac{n!}{r!(n-r)!}\\&=\binom{n}{r}.\end{align}$$
