Combinations Equation to prove (possibly with Pascal's relation?) Can this equation be proved?
$${n \choose p} = {n-2 \choose p} + 2 {n-2 \choose p-1} + {n-2 \choose p-2}$$
Here is what I've tried:
$${ 
   {n!\over (n - p)! p!} = {(n - 2)! \over (n - p -2)! p!} + {2(n - 2)!\over (n-p-1)! (p - 1)!} + {(n-2)!\over(n-p)!(p-2)!}
   }$$
Then trying to get the common denominator:
$${
   {n!\over (n - p)! p!} = {(n - 2)! \over (n-p-2)!p(p-1)(p-2)!} + {2(n - 2)!\over (n-p-1)(n-p-2)!(p-1)(p-2)!} + {(n - 2)! \over (n-p)(n-p-1)(n-p-2)!(p-2)!}
   }$$
Then:
$${
{n!} = (n-p)(n-p-1)(n-2)!+2p(n-p)(n-2)!+p(p-1)(n-2)!
}$$
factorizing this gets me:
$${
{n!} = (n-2)![(n-p)(n-3p-1)+p(p-1)]
}$$
and that's it.
 A: Combinatorial interpretation:
On the left side, you have all possibilities to choose $p$ elements from set $\{1,2,...,n\}$. 
On the right side it is the same because.
You can do a set without $n, n-1$. So you have $\{1,...,n-2\}$.
Now you have two sets $\{n,n-1\}$ and $\{1,...,n-2\}$ and trying to combine elements from two of them to have all $p$ - element subsets of $\{1,2,...,n\}$
1) You choose $p$ elements from set $\{1,...,n-2\}$, so $\binom{n-2}{p}$
2) You choose $p-1$ elements from this set and ADD in TWO possible ways ONE element from $\{n-1,n\}$ so you have $2\binom{n-2}{p-1}$ 
3) You choose $p-2$ elements from this set and have no other option except to take both $\{n-1,n\}$ to have a set of $p$ elements so $\binom{n-2}{p-2}$.
Cases 1), 2), 3) represent all the possibilities, so you add them and have right side.
A: The identity was proved using Pascal's Rule twice as suggested by @Raskolnikov. Thanks All.
A: You want an algebraic proof, here it comes:
Actually you did everything alright, but you made one mistake, it is 
\begin{align}
{n \choose p} &= {n-2 \choose p} + 2 {n-2 \choose p-1} + {n-2 \choose p-2}
\end{align}
which gives you after factorizing (everything good until here)
$$
n! = (n-p)(n-p-1)(n-2)!+2p(n-p)(n-2)!+p(p-1)(n-2)!
$$
Now here comes the mistake, it is not $(n-3p-1)$ but
\begin{align}
n! &= (n-p)(n-p-1)(n-2)!+2p(n-p)(n-2)!+p(p-1)(n-2)!\\
&=(n-2)!\left[(n-p)(n-p-1+2p)+p(p-1)\right]\\
&=(n-2)!\left[(n-p)(n\color{red}{+p-1})+p(p-1)\right]
\end{align}
which goes into
$$
\Leftrightarrow (n-1)n=n^2-n=\left[(n-p)(n+{p-1})+p(p-1)\right]
$$
and this leads you directly to an equality, the $p$'s are cancelling out and you got what you were looking for!
