Full of $\binom{\text{summations}}{\text{combinations}}$

I will use the Symmetry Identity a lot in this proof, so here it is for reference:

If $b \ge a \ge 0$ , then $\binom {b} {a} = \binom {b} {b-a}$

Let's first simplify the first sum:

$\sum_{k=1}^{333} \binom {k+2013} {k+1565} = \sum_{k=1}^{333} \binom {k+2013} {448} = \sum_{x=2014}^{2346} \binom {x} {448} = \sum_{x=448}^{2346} \binom {x} {448} - \sum_{x=448}^{2013} \binom {x} {448}$

Now we use the "Hockey Stick Identity" which says that for $b \ge a \ge 0$ , we have $\sum_{x=a}^{b} \binom {x} {a} = \binom {b+1} {a+1}$ I won't prove it here, but it's easy to prove by induction.

Applying this formula to the above gives us: $\sum_{k=1}^{333} \binom {k+2013} {k+1565} = \binom {2347} {449} - \binom {2014} {449}$

We use the Hockey Stick Identity again to simplify the second sum: $\sum_{k=1}^{48} \binom {k+2346} {1897} = \sum_{x=2347}^{2394} \binom {x} {1897} = \sum_{x=1897}^{2394} \binom {x} {1897} - \sum_{x=1897}^{2346} \binom {x} {1897} = \binom {2395} {1898} - \binom {2347} {1898}$

Now we put everything together, and cancel things out:

$\binom {2014} {1565} + \left( \binom {2347} {449} - \binom {2014} {449} \right) + \left( \binom {2395} {1898} - \binom {2347} {1898} \right)$

$= \binom {2014} {1565} + \binom {2347} {1898} - \binom {2014} {1565} + \binom {2395} {1898} - \binom {2347} {1898}$

$= \binom {2395} {1898}$

Thus, $x - y = 497$