This is a workshop task i've been given for my combinatorics module, the only hint i've been given towards it is: 'Suppose we have n blue balls and n red balls. What do both quantities above describe?'
I started thinking about the problem by noting that the right hand side is the number of subsets of $2n$ of size $n$. And also $\binom{n}{i}$ is the number of subsets of n of size $i$. so let $i=0$ and we have $1$ way to make a subset of size zero, then $\binom{n}{1} = n$ and so on. Thinking about it with the 'hint' i can describe the LHS as being the number of different ways we can take $i$ balls out of the bag and when all combinations are added together, it becomes the total number of ways we can take any number of balls ($i \leq n$) from the bag.
The RHS is the number of subsets of $2n$ of size $n$, or the number of ways we can take $n$ balls out of a bag of size $n$, using the hint, if we have a bag of $n$ blue and $n$ red balls it's the total number of ways we can pull out $n$ number of balls where so many are red and blue, e.g $1$ red and $(n-1)$ blue.
I can see all this but I have no clue how either side even relates let alone is equal.
$$(1+x)^n(1+x)^n=(1+x)^{2n}.$$
$$(1+x)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^k.$$
$$(1+x)^n(1+x)^n=\sum_{i=0}^n\sum_{j=0}^n\binom{n}{i}\binom{n}{j}x^{i+j}.$$ Therefore, $$\binom{2n}{n}=\sum_{i=0}^n\sum_{i+j=n}\binom{n}{i}\binom{n}{j}=\sum_{i=0}^n\binom{n}{i}\binom{n}{n-i}=\sum_{i=0}^n\binom{n}{i}^2.$$