Algebraic indices in the sigma symbol

Question

$ \sum\limits_{x + y=0}^n 1/(x!.y!)$

for example let n = 3.

Then should we right down all the combinations then sum them does this means that ?:

$ \sum\limits_{x + y=0}^3 1/(x!.y!) = $

$ x + y = 0 $ 1 case => 1/0!.0!

$ x + y = 1 $ 2 cases => 1/1!.0! + 1/0.1!

$ x + y = 2 $ 3 cases => 1/1!.1! + 1/2.0! + 1/0!.2!

$ x + y = 3 $ 4 cases => 1/0!.3! + 1/1.2! + 1/2!.1! + 1/3!.0!
...

Answer 1

Yes, that's correct. ${}{}{}{}{}{}$