Q:

Handling summations with two variables

If I have a summation with let's say $x=0 \dots 500$ and $y=0\dots1500$ 
$500 \choose x$ $ 1500 \choose y$ $\dfrac{1}{2^{500}}\dfrac{2^{1500-y}}{3^{1500}}$, 
How would I handle the constant? If I split the summations apart into two separate one's would I include it in both? And when splitting them, would I add or multiply the two summations?

A:

Assuming that you mean $$\sum_{x=0}^{500}\sum_{y=0}^{1500}\binom{500}x\binom{1500}y\frac1{2^{500}}\frac{2^{1500-y}}{3^{1500}},$$ there are a few useful things to observe. Since the sum is finite, we can switch the order to $$\sum_{y=0}^{1500}\sum_{x=0}^{500}\binom{500}x\binom{1500}y\frac1{2^{500}}\frac{2^{1500-y}}{3^{1500}}.$$ Now, we can pull constant multiples out to the front (because of distributivity), giving us $$\frac1{2^{500}3^{1500}}\sum_{y=0}^{1500}\sum_{x=0}^{500}\binom{500}x\binom{1500}y2^{1500-y}.$$ Now, again by distributivity, we can factor out the sum of the $x$ terms, giving us $$\frac1{2^{500}3^{1500}}\left[\sum_{x=0}^{500}\binom{500}x\right]\left[\sum_{y=0}^{1500}\binom{1500}y2^{1500-y}\right].$$
The sum of the $x$ terms is $2^{500},$ giving us $$\frac1{3^{1500}}\sum_{y=0}^{1500}\binom{1500}y2^{1500-y}$$ after cancellation. Now we're down to one sum, and I leave the rest to you.

Edit: You've discovered the trick of using binomial expansion to get the last sum done (excellent work). I'll add that here in the answer, so that future users might see how it was done. Hover over the white space below to see it.

Since $(a+b)^{1500}=\sum\limits_{y=0}^{1500}\binom{1500}ya^yb^{1500-y},$ then using $a=1$ and $b=2$ shows us that the final sum is $1$.

