The Marginal Distribution of a Multinomial The binomial distribution is generalized by the multinomial distribution, which follows:

 \begin{align}
f(x_1,\ldots,x_k;n,p_1,\ldots,p_k) & {} = \Pr(X_1 = x_1\mbox{ and }\dots\mbox{ and }X_k = x_k) \\  \\
& {} = \begin{cases} { \displaystyle {n! \over x_1!\cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}}, \quad &
\mbox{when } \sum_{i=1}^k x_i=n \\  \\
0 & \mbox{otherwise,} \end{cases}
\end{align}


In particular, the "three"nomial distribution follows:
$${n! \over x_1! x_2!(n-x_1-x_2)!}p_1^{x_1}p_2^{x_2}p_3^{n-x_1-x_2}$$
I am not able to show why the marginal probability of this distribution, with respect to either $x_1$ or $x_2$ follows $b(n, p_1)$ or $b(n, p_2)$, respectively.
Please help!
 A: The simplest way is "combinatorial/probabilistic." Recall that the "three-nomial" distribution measures the probability of $x_1$ Type 1 events, $x_2$  Type 2 events, and $n-x_1-x_2$ Type 3 events, when an experiment is repeated independently $n$ times, with probabilities of "success" respectively equal to $p_1$, $p_2$, and $p_3$, where $p_1+p_2+p_3=1$. 
The probability of $x_1$ Type 1 events is therefore 
$$\binom{n}{x_1}p_1^{x_1}(p_2+p_3)^{n-x_1}.\tag{1}$$
It follows that the marginal distribution of $X_1$ is binomial. 
If we really wish to sum, by the Binomial Theorem the probability (1) is equal to
$$\binom{n}{x_1}p_1^{x_1}\sum_{x_2=0}^{n-x_1}\binom{n-x_1}{x_2}p_2^{x_2}p_3^{n-x_1-x_2}.$$
This is precisely the same as the result of summing over all (appropriate) $x_2$ the probability that $X_2=x_2$ and $X_3=n-x_1-x_2$. If we want to save writing down (1) until the very end, we can just write the sum argument backwards.
A: To see it is relatively easy by the sum of indicators approach.
To verify/proof from the trinomial case:
$$\begin{align} 
\Pr\{X_1 = x_1\} &= \sum_{x_2=0}^{n-x_1} \Pr\{X_1 = x_1, X_2 = x_2\} \\
&= \sum_{x_2=0}^{n-x_1} \frac {n!} {x_1!x_2!(n-x_1-x_2)!}
p_1^{x_1}p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2}\\
&= \frac {n!} {x_1!}p_1^{x_1}  \sum_{x_2=0}^{n-x_1}
\frac {1} {x_2!(n-x_1-x_2)!} p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2} \\
&= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1}  \sum_{x_2=0}^{n-x_1}
\frac {(n - x_1)!} {x_2!(n-x_1-x_2)!} p_2^{x_2}(1-p_1-p_2)^{n-x_1-x_2} \\
&= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} [p_2 + (1 - p_1 - p_2)]^{n-x_1} \\
&= \frac {n!} {x_1!(n-x_1)!}p_1^{x_1} (1 - p_1)^{n-x_1}, x_1 = 0, 1, \ldots, n 
\end{align}$$
So the only key steps are: recognizing the support when $X_1 = x_1$ and recognizing the binomial theorem / expansion. 
