Using joint probability mass functions (multiple parts)

This question has a lot to it, and I'm guessing it has to be done in order (if so, you can't skip ahead a part without completing the prior part). The question(s) are listed below word-for-word.

Let $\displaystyle X$ and $\displaystyle Y$ have the following pmf:

$\displaystyle P(X=i,Y=j)=\frac{\alpha}{(1+i+j)!}, i\geq 0, j\geq 0$

for some $\displaystyle \alpha>0$.

- Explain without any calculation why $\displaystyle X$ and $\displaystyle Y$ have the same marginal pmf. That is, why $\displaystyle P(X=i)=P(Y=i)$.
- Let $\displaystyle S=X+Y$. Show that $\displaystyle P(S=k)=\frac{\alpha}{k!}$ for all $\displaystyle k\geq0$. Note that $\displaystyle X$ and $\displaystyle Y$ need not be independent (as will be discussed below).
- Conclude the value of $\displaystyle \alpha$ and recognize the distribution of $\displaystyle S$. What is the parameter of this distribution?
- Compute $\displaystyle P(X=0)$. Are $\displaystyle X$ and $\displaystyle Y$ independent?
- Find $\displaystyle E(S)$, and conclude $\displaystyle E(X)$.
**HINT**: To find $\displaystyle E(X)$ from $\displaystyle E(S)$ you may use part (a) that says $\displaystyle X$ and $\displaystyle Y$ have same distributions and therefore same... - Compute $\displaystyle P(X=Y)$.
**HINT**: You can write it as

$\displaystyle P(X=Y)=\sum_{i=0}^\infty P(X=i,Y=i)$,

and think of the infinite expansion of $\displaystyle (e^x-e^{-x})/2$. - Conclude $\displaystyle P(X>Y)$.
**HINT**: You may use a symmetry argument.

This is a lot to do, but that's the question as a whole, and I'd rather not make multiple topics pertaining to the same joint pmf.