Is there a simple way to prove this identity using double summation? I am trying to prove that $$\sum_{k=0}^n\sum_{r=0}^k{n \choose k}{k \choose r}(-1)^{n+k}A_r=\sum_{r=0}^n\sum_{k=r}^n{n \choose k}{k \choose r}(-1)^{n+k}A_r$$
By writing out the full sum I can see that this is true, but is there a method based on purely some properties of double summation?
 A: You have stumbled upon a discrete version of Fubini's theorem! We do indeed have for any two-index sequence $a_{k,r}$,
$$ \sum_{k=0}^n \sum_{r=0}^k a_{k,r} = \sum_{\left\{(k,r) ∈ \Bbb Z^2 : \substack{0\leq k \leq n \\ 0 \leq r \leq k } \right\}} a_{k,r}$$
And the set we are summing over can be rewritten,
$$ \left\{(k,r) ∈ \Bbb Z^2 : \substack{0\leq k \leq n \\ 0 \leq r \leq k } \right\}=\left\{(k,r) ∈ \Bbb Z^2 : \substack{0\leq r \leq n \\ k \leq r \leq n } \right\}$$
and then we can rewrite
$$\sum_{k=0}^n \sum_{r=0}^k a_{k,r} = \sum_{\left\{(k,r) ∈ \Bbb Z^2 : \substack{0\leq r \leq n \\ k \leq r \leq n } \right\}} a_{k,r} = \sum_{r=0}^n \sum_{k=r}^n a_{k,r}$$
Here is a badly drawn diagram: The set above is the set of integer-points in the following triangle: 

A: The nature of the summand doesn't actually matter here: one can show that
$$ \sum_{k=0}^n \sum_{r=0}^k a_{k,r} = \sum_{r=0}^n \sum_{k=r}^n a_{k,r}. \tag{1} $$
Why? We're not summing over the whole square $0\leqslant k,r\leqslant n$, but instead, the just the triangle where $r \leqslant k$:
$$ \sum_{ \substack{0 \leqslant k,r \leqslant n \\ r \leqslant k }} $$
This means that if we swap the order of summation (and sum "vertically" instead of "horizontally", we have to start $k$ at $r$ and sum up to $n$, then do the sum over $r$, so you end up with (1).
