Find the coefficient of $w^{3}x^{5}z^{2}$ in the expansion of $(w+x+y+z)^{10}$ I need to find the coefficient of $w^{3}x^{5}z^{2}$ in the expansion of $(w+x+y+z)^{10}$.
How can I do this?
The solution is probably connected somehow to ordinary generating functions. I found something similar:
coefficient of $x^{17}$ in $(1+x^{5}+x^{7})^{20}$ is $C^{20}_{2}*18 = 3420$...but I have no idea why, and what does $C^{20}_{2}$ mean.
 A: Firstly, to address matters of notation,
$$ C^{20}_2 = \binom{20}{2} = \frac{20!}{18! \times 2!}$$
Now, to address the coefficient of $w^3x^5z^2$, this is effectively the same as "picking" $w$ $3$ times, $x$ $5$ times and $z$ twice from the brackets in the expansion. Thus, if we start with the $w$'s, we need to choose $w$ $3$ times from $10$ brackets, or $C^{10}_3$ times. From the remaining $7$ brackets, we can then pick $x$ $5$ times, which we can do $C^7_5$ times, and then the remaining terms must be $w$'s, so the total ways we can "choose" $w^3x^5z^2$ is 
$$  C^{10}_3 \times C^{7}_5 = \frac{10!}{7! \times3!} \times \frac{7!}{5! \times 2!} = \frac{10!}{3! \times 5! \times 2!} = \binom{10}{3,5,2} = 2520$$
where the symbol at the end is the multinomial coefficient. This type of argument leads to a proof of the multinomial theorem mentioned by Abel.
A: When we expand $$(w+x+y+z)^{10}$$ we obtain a sum of degree $10$ monomials.
We get a $+1$ contribution to the coefficient of $w^3x^5z^2$ from the coefficients highlighted below:
$$(\color{blue}{\bf w}+x+y+z)(\color{blue}{\bf w}+x+y+z)(\color{blue}{\bf w}+x+y+z)(w+\color{red}{\bf x}+y+z)(w+\color{red}{\bf x}+y+z) \times\\
  (w+\color{red}{\bf x}+y+z)(w+\color{red}{\bf x}+y+z)(w+\color{red}{\bf x}+y+z)(w+x+y+\color{green}{\bf z})(w+x+y+\color{green}{\bf z})$$
$$(w+\color{red}{\bf x}+y+z)(w+x+y+\color{green}{\bf z})(\color{blue}{\bf w}+x+y+z)(w+\color{red}{\bf x}+y+z)(w+\color{red}{\bf x}+y+z) \times\\
  (\color{blue}{\bf w}+x+y+z)(\color{blue}{\bf w}+x+y+z)(w+\color{red}{\bf x}+y+z)(w+x+y+\color{green}{\bf z})(w+\color{red}{\bf x}+y+z)$$
and so on.
If we write down the highlighted terms as a list, we get $$wwwxxxxxzz$$ and $$xzwxxwwxzx$$ for the two cases above.
So, the contributions to the coefficient of $w^3x^5z^2$ correspond to orderings (or multiset permutations) of the multiset $$\{w,w,w,x,x,x,x,x,z,z\}.$$  There are $$\binom{10}{3,5,2}$$ such orderings; this is counted by the multinomial coefficient.
A: According to the Multinomial Theorem the answer is ${10\choose 3,5,2}= \frac{10!}{3!5!2!} = 2520$.
