= Solution
The <independence> of $X$ and $Y$ permits factoring the <expectation> of the product $e^{tX}e^{tY}$. The <moment-generating function of a binomial distribution> from the previous part therefore gives
$$
\begin{aligned}
M_{X+Y}(t)&=\mathbb E[e^{t(X+Y)}]
=\mathbb E[e^{tX}]\mathbb E[e^{tY}]\\
&=(1-p+pe^t)^n(1-p+pe^t)^m
=\boxed{(1-p+pe^t)^{n+m}}.
\end{aligned}
$$
This is the <moment-generating function> of $B(n+m,p)$. Both <moment-generating functions> are finite on a neighbourhood of zero, so the <uniqueness theorem for moment-generating functions> identifies the <probability distribution>:
$$
\boxed{X+Y\sim B(n+m,p).}
$$
The common success <probability> $p$ and <independence> are both used in this argument.
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