Computing $C_0^2+C_1^2+C_2^2+C_3^2+ \cdots +C_n^2$ If $C_k$ denotes binomial coefficient of choosing $k$ objects from  a set of $n$ objects how to calculate this:

$$C_0^2+C_1^2+C_2^2+C_3^2+\cdots +C_n^2$$
Consider a group of $2n$ people of which $n$ are men and $n$ are women. In how many ways can we choose $n$ people from this group? In $\binom{2n}{n}$ ways.  Now we count the same number in a different way. Each such chosen group consists of $k$ men and $n-k$ women for some $k=0,1,2,...,n$. First we choose $k$ men from $n$ men; this can be done in $\binom{n}{k}$ ways. Then select $n-k$ women from $n$ women; this can be done in $\binom{n}{n-k}$ ways. So, number of ways of choosing $n$ people of which $k$ are men and $n-k$ are women is $\binom{n}{k} \binom{n}{n-k}$. And $k$ ranges from $0,1,2,...,n$. Therefore, the total number of ways of choosing $n$ people is $\sum_{k=0}^n{\binom{n}{k} \binom{n}{n-k}}$. Thus we get, $$ \binom{2n}{n} = \sum_{k=0}^n{\binom{n}{k} \binom{n}{n-k}} $$
Now use the identity $\binom{n}{k} = \binom{n}{n-k}$. [To see this, observe that number of ways of choosing $k$ objects from $n$ objects is the same as discarding $n-k$ objects from $n$ objects.] We get, $$ \binom{2n}{n} = \sum_{k=0}^n{\binom{n}{k} \binom{n}{k}} $$
$$ \binom{2n}{n} = \sum_{k=0}^n{{\binom{n}{k}^2 }} $$
From Wikipedia
$$\sum_{k=0}^n {n \choose k}^2 = {2n\choose n}.$$