Find the coefficient of $w^{3}x^{5}z^{2}$ in the expansion of $(w+x+y+z)^{10}$

Question

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.

I don't see how its possible to find the coefficient of $w^3x^5z^2$ unless you mean $w^3x^5z^2y^{10}$. — Andrew D, Sep 05 '13 at 14:22
I'm sorry, that's $(w+z+y+z)^{10}$, not $(w+z+y+z)^{20}$ It should be correct now. — khernik, Sep 05 '13 at 14:24

score 8 · Accepted Answer · answered Sep 05 '13 at 14:39

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.

Nice job. You beat me to the answer by less than a minute . – Rick Decker Sep 05 '13 at 14:42 — Rick Decker, Sep 05 '13 at 14:42

score 8 · Answer 2 · answered Sep 05 '13 at 14:42

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.

score 7 · Answer 3 · answered Sep 05 '13 at 14:29

7

According to the Multinomial Theorem the answer is ${10\choose 3,5,2}= \frac{10!}{3!5!2!} = 2520$.

answered Sep 05 '13 at 14:29

Abel

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Find the coefficient of $w^{3}x^{5}z^{2}$ in the expansion of $(w+x+y+z)^{10}$

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