Let $m$ and $n$ be positive integers. Then Multinomial Theorem states that $$(x_{1}+x_{2}+\cdots +x_{m})^{n}=\displaystyle\sum_{\begin{array}{c}\\k_{1}+\cdots +k_{m}=n\\ 0\leq k_{j},\ 1\leq j \leq m\end{array}} \frac{n!}{k_{1}!k_{2}!\cdots k_{m}!}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}$$
This theorem is the generalization of Binomial Theorem.
Binomial Theorem states that the coefficient of $x_{1}^{n-k}x_{2}^{k}$ when we expand $(x_{1}+x_{2})^{n}$ is given by $x_{1}^{n-k}x_{2}^{k}$.
On the other side, Multinomial Theorem states that the coefficient of $x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}$ when we expand $(x_{1}+x_{2}+\cdots +x_{m})^{n}$ is given by $\frac{n!}{k_{1}!k_{2}!\cdots k_{m}!}$.
For example, if we expand $(x_{1}+x_{2}+x_{3})^{3}$, the coefficient of $x_{1}^{2}x_{2}^{1}$ is given by $\frac{3!}{2!1!0!}=3$. ( Note that $0!=1$ )
Similar to Binomial coefficient, we sometimes denote Multinomial coefficient as below：
$$\frac{n!}{k_{1}!k_{2}!\cdots k_{m}!}=\left(\begin{array}{ccc} \ &n & \\k_{1}&\cdots & k_{m}\end{array}\right)$$
Examples
Find the coefficient of $x^{2}y^{2}z^{3}$ when $(x+2y+5z)^{7}$ is expanded.
Solution：The general term of $(x+2y+5z)^{7}$ is given by
$$\frac{7!}{k_{1}!k_{2}!k_{3}!}x^{k_{1}}(2y)^{k_{2}}(5z)^{k_{3}}=\left(\frac{7!}{k_{1}!k_{2}!k_{3}!}2^{k_{2}}5^{k_{3}}\right)x^{k_{1}}y^{k_{2}}z^{k_{3}}$$
Thus, if we set $k_{1}=2,\ k_{2}=2,\ k_{3}=3$, the coefficient of $x^{2}y^{2}z^{3}$ is obtained as below：
$$\frac{7!}{2!2!3!}2^{2}5^{3}=105000$$
Find the coefficient of $x^{3}$ when $(1+x+2x^{2})^{5}$ is expanded.
Solution：The general term of $(1+x+2x^{2})^{5}$ is given by
$$\frac{5!}{k_{1}!k_{2}!k_{3}!}1^{k_{1}}x^{k_{2}}(2x^{2})^{k_{3}}=\left(\frac{5!}{k_{1}!k_{2}!k_{3}!}2^{k_{3}}\right)x^{k_{2}+2k_{3}}$$
and the term $x^{3}$ is obtained when $k_{2}+2k_{3}=3$. Thus there are two possible cases such as $(k_{2}, k_{3})=(1,1),\ (3,0)$ .
At this time, since $k_{1}+k_{2}+k_{3}=5$, each combinations are given by $(k_{1}, k_{2},k_{3})=(3,1,1),\ (2,3,0)$ . Therefore, the coefficient of $x^{3}$ is given by
$$ \frac{5!}{3!1!1!}2^{1}+\frac{5!}{2!3!0!}2^{0}=50$$