Let $X$ and $Y$ be discrete random variables with joint probability function
$$p_{XY} (x, y)=\begin{cases} {n!y^x(pe^{-1})^y(1-p)^{n-y}}\over{y!(n-y)!x!}, & y = 0, 1, · · · , n\ ; x = 0, 1, · · · \\ 0, & \text{otherwise} \end{cases}$$
Find $p_Y (y)$.
It seems like this is the binomial distribution multiplied by the poisson distribution, and indeed the solution for $p_Y (y)$ is the binomial distribution of $y$.
However, I would like to know how to arrive at this solution using the formula $$p_Y (y)={\sum_{x}}p_{XY} (x, y)$$
as it looks very complicated in this particular distribution. Is there a simple way to do it?
It's not difficult. You only need to sum the terms that depend on $x$, the rest go outside the summation. Also, recall the exponential series:
$$ \sum_{x=0}^\infty \frac{y^x}{x!}=e^y$$
and you'll get the Binomial.