Combinatorial proof of ${{n+1}\choose 3}-{{n-1}\choose 3}=(n-1)^2.$ 
Prove that ${{n+1}\choose 3}-{{n-1}\choose 3}=(n-1)^2.$

I found the algebraic proof of the above statement.
So we have to show that $$ \frac{(n+1)(n)(n-1)}{3\times 2}-\frac{(n-1)(n-2)(n-3)}{3\times 2} \stackrel{?}{=} (n-1)(n-1)$$ $$\implies \frac{(n+1)(n)}{3\times 2}-\frac{(n-2)(n-3)}{3\times 2} \stackrel{?}{=} (n-1) $$$$ \implies (n+1)n-(n-2)(n-3)\stackrel{?}{=}3 \times 2 
(n-1), $$ which is true and we are done!
But I couldn't get the combinatorial proof. Any hints?

I would use Pascal's identity to get
$$\binom{n+1}{3}-\binom{n-1}{3} = \left[\binom{n+1}{3} - \binom{n}{3}\right] + \left[\binom{n}{3}-\binom{n-1}{3}\right] = \binom{n}{2} + \binom{n-1}{2}.$$
This can be solved algebraically:
$$\binom{n}{2} + \binom{n-1}{2} = \frac{n(n-1)}{2} + \frac{(n-1)(n-2)}{2} = \frac{(n-1)[(n-2)+n]}{2} = (n-1)^{2}.$$
I'm shocked I figured this out so fast. Sometimes you just see the solutions when you read the problem and sometimes you get stuck on a wrong path. It's just the story of math :)
