Finding coefficient of $j^2k^3lm^3$ in $(j + k + l + m)^9$

Question

I'm trying to find the coefficient of:

$$j^2k^3lm^3$$

in:

$$(j + k + l + m)^9$$

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According to the Book of Proof (which is our material), it seems for:

$$x^ay^b, \{a,b\} \in \mathbb{N}$$

in:

$$(x+y)^c, c \in \mathbb{N}, c \geq \{a,b\}$$

we can use the binomial theorem:

$$(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + ... + \binom{n}{n-1}xy^{n-1} + \binom{n}{n}y^n$$

However, I'm lost on the other one. Please help?

Answer 1

In general, the coefficient of $x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_k^{\alpha_k}$ (with $\alpha_1 + \alpha_2 + \cdots + \alpha_k = n$) in the expansion of $(x_1 + x_2 + \cdots + x_k)^n$ is given by the multinomial $$\binom{n}{\alpha_1, \: \alpha_2, \: \dots, \: \alpha_k} = \frac{n!}{\alpha_1!\alpha_2!\cdots\alpha_k!}$$

In your case the coefficient of $j^2k^3lm^3$ is $$\binom{9}{2, \: 3, \: 1, \: 3} = \frac{9!}{2!3!1!3!} = 5040$$