I am struggling to show the following are equal. I must be missing a simple piece somewhere but just cant spot it. I want to let the LHS=RHS
$k(k-1) $$n\choose k$$= n(n-1)$ $n-2\choose k-2$
I can get the LHS down to $(k-1)n!/(k-1)!(n-k)!$ I cant get the RHS near this..
When one talks about a "combinatorial proof" one generally invokes the principle of double counting: if one can count how many outcomes a scenario has correctly in more than one way, then the expressions used for the number of outcomes must be equal.
Here, we have the hypothetical scenario of having $n$ people in a committee. We want to count how many subcommittees can be formed with a specific chairman and a specific vice-chairman of size $k$ (including the chairman and vice chairman).
Method one:
First pick who the $k$ people for the subcommittee are from the $n$ people. This can be accomplished in $\binom{n}{k}$ ways.
Now, from those $k$ people that we chose, choose one of them to act as the chairman. This can be accomplished in $k$ ways.
Then choose another to act as the vice chairman from the remaining $k-1$ people in the subcommittee who hadn't been selected to be chairman. This can be accomplished in $k-1$ ways.
The total count can be arrived at via multiplication principle to be $\binom{n}{k}k(k-1)$
Method two:
First pick who will be the chairman for the subcommittee from the $n$ people. This can be accomplished in $n$ ways.
Then pick who will be the vice chairman for the subcommittee from the $n-1$ people remaining. This can be accomplished in $n-1$ ways.
Finally, pick the remaining $k-2$ followers for the subcommittee from the $n-2$ remaining people. This can be accomplished in $\binom{n-2}{k-2}$ ways.
by multiplication principle this yields a total of $\binom{n-2}{k-2}n(n-1)$
By the first principle of counting, we have then that the two expressions must be equal, i.e. $\binom{n}{k}k(k-1)=\binom{n-2}{k-2}n(n-1)$
If you are looking for an algebraic proof (which is often seen as tedious, requiring less understanding, and generally inelegant), then note:
$\binom{n}{k}k(k-1)=\frac{n!k(k-1)}{k!(n-k)!}=\frac{n!}{(k-2)!(n-k)!}=\frac{n(n-1)(n-2)!}{(k-2)!((n-2)-(k-2))!}=\binom{n-2}{k-2}n(n-1)$
The only things really needing to be noted are the algebraic representation of the binomial coefficient $\binom{a}{b}=\frac{a!}{b!(a-b)!}$ and the properties of the factorial $a!=a(a-1)!$