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.