Why sum from $0 \leq k \leq n$? Combinatorial Proof of $\sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$ It's subsumed under Vandermonde's Identity.  

Let a set consist of $n$ kiwis and $n$ figs. We want to count how many ways there are to select $n$ fruits, regardless of type. One immediate answer is $\binom{2n}{n}$. 
Now count in a different way. Each selection of $n$ fruits contains a certain number, say $k$, of kiwis and also $n-k$ figs, so $0 \leq k \leq n$.
  For any particular value of $k$, we can choose the kiwis in $\binom{n}{k}$ ways and the figs in $\binom{n}{n - k}$ ways. 
So the #
  of ways to choose an $n$ set with $k$ kiwis is $\binom{n}{k}\binom{n}{n - k} = \binom{n}{k}^2$. 



*

*I initially asked this here, but now separately.



Thus, $\color{green}{\text{the total # of ways to select $n$ fruits independent of type}}$
  = total # of ways to select $0$ kiwis + total # of ways to select $1$ kiwis $+ \cdots +$ total # of ways to select $(n - 1)$ kiwis + $\color{orangered}{\text{the total # of ways to select $n$ KIWIS}}$ $\color{green}{= \binom{n}{0}^2 + ... \binom{n}{n - 1}^2 + \color{orangered}{\binom{n}{n}^2}}$. $\Large{\color{#B22222}{]}}$



*Why must we sum from $0 \leq k \leq n$?

 A: If you have chosen $k$ kiwis, then you need to choose $n-k$ figs.  The number of ways of choosing $n-k$ figs from $n$ figs is ${n \choose n-k}$ which is equal to ${n \choose k}$.  Multiply this by the number of ways of choosing $k$ kiwis from $n$ and you get the number of ways of choosing $k$ kiwis and $n-k$ figs.
To get all possibilities, you have to consider all possibilities, ranging from choosing no kiwis to choosing all kiwis, i.e. choosing from $k=0$ to $k=n$ kiwis, so you need to sum over this range. 
A: *

*No, picking exactly $k$ kiwis means you discount the $n-k$ remaining kiwis, but you still have to chose $n-k$ figs out of $n$, which is the second factor.  

*You can pick anywhere from $0$ (none) to $n$ (all) kiwis. $n+1$ is not allowed ($=0$ possibilities) and "adding" a kiwi (choosing $-1$ ~ adding a $n+1$-st Kiwi to the stack) is also disallowed.

A: Since there are ${2n\choose n}$ ways to select $n$ pieces of fruit regardless of the type we can select $k$ of one type of fruit and the remaining $n-k$ of the other type of fruit. We then sum for all possible $k$ values from $0\leq k\leq n$ to obtain every possible combination.
