How does: $n\binom{n-1}{r-1} = r\binom{n}{r}$ I'm trying to figure out the identity above though I'm having difficulties towards figuring it out and would kindly appreciate your support!
$n\binom{n-1}{r-1} = r\binom{n}{r}$
What I have tried:
Given that
$$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$$
Then by rearranging for $\binom{n-1}{r-1}$ I get
$$\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}$$
Which simplifys to:
$$n\left[\frac{n!}{(n-r)!r!}-\frac{(n-1)!}{(n-r!)r!}\right]=n\cdot\frac{n!-(n-1)!}{(n-r!)r!}$$
I'm stuck here on how to simply this any further to get the result I'm after.
 A: $$\frac nr \binom{n-1}{r-1} = \frac nr \frac{(n-1)!}{(r-1)! (n-r)!} = \frac{n!}{r!(n-r)!} = \binom nr$$
A: Of course this just falls out from the identity $\begin{pmatrix}p\\q\end{pmatrix}=\frac{p!}{(p-q)!q!}$. That's no fun (also unenlightening). Outline of a "combinatorial" proof:
Say $S=\{1,2,\dots,n\}$. Let $X$ be the set of all ordered pairs $(s,E)$ such that $s\in S$, $E\subset S$, $E$ has $r-1$ elements, and $s\notin E$. The identity follows by counting the number of elements of $X$ in two different ways.
A: $\binom{n - 1}{r - 1} = \frac{(n - 1)!}{(r - 1)!(n - r)!}$.
$n*\frac{(n - 1)!}{(r - 1)!(n - r)!} = \frac{n!}{(r - 1)!(n - r)!}.$
Dividing the LHS and RHS by $r$, we want to prove that $\frac{n!}{(r - 1)!(n - r)!(r)} = \binom{n}{r}$.
Expanding the RHS, we get $\binom{n}{r} = \frac{n!}{r!(n - r)!}$
Simplifying the LHS, we get $\frac{n!}{r!(n - r)!}$.
RHS = $\frac{n!}{r!(n - r)!}$ = LHS.
So, we are done.
A: From the formula for binomial coefficients, but the other way: what has to be proved is
$$ \binom nr=\frac nr \binom{n-1}{r-1}.$$
Now, we have
$$\binom nr=\frac{n!}{r!(n-r)!}=\frac nr\frac{(n-1)!}{(r-1)!(n-r)!}=\frac nr\binom{n-1}{r-1}$$
since $\: n-r=(n-1)-(r-1)$.
