prove the combinatorial identity: $ {2n \choose n} + {2n\choose n-1} = (1/2){2n + 2 \choose n + 1}$ So the first thing I did was to multiply and then isolate the left-most term. 
That leaves me with  $$ 2{2n \choose n} ={2n + 2\choose n + 1} - 2{2n \choose n -1} $$
No specific reasoning except that it looks cleaner, at least to me. Then the way I thought about it was that we are trying to make two $n$ member committees out of $n$ koalas and $n$ pandas. I was told when seeing subtraction I should think of complement but I don't really know how to incorporate the complement into this. Any way, am I on the right path or completely off. Thank you in advance for your time. 

Note that $\binom {2n}{n-1}=\binom {2n}{n+1}$. So we can prove 
$$\binom {2n+2}{n+1}=\binom {2n}{n+1}+2\binom {2n}{n}+\binom {2n}{n-1}.$$
Now you can prove this by considering the number of ways to choose a committee of size $n+1$ from a group of $2n+2$. Suppose there are two people named A and B in the group. For the RHS, you need to count separately based on A and/or B being in the committee or not. This is a special case of Proving the combinatorial identity ${n \choose k} = {n-2\choose k-2} + 2{n-2\choose k-1} + {n-2\choose k}$.
