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

I'm trying to find the coefficient of:

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

in:

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

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:

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

• Well, there is the multinomial theorem, which is like the binomial theorem, only... multier. Alternatively, use the binomial theorem to find the coefficient of $j^2(k+l+m)^7$ in $(j+(k+l+m))^9$, and then continue inwards. – Arthur Sep 25 '17 at 13:40
• Tried to write it down as $\big( (j+k) + (l+m) \big)^9$ the use binomial where $x=(j+k),y=(l+m)$ then u can expand the bionomial inside the terms of the previous binomial and then u can – Pookaros Sep 25 '17 at 13:44
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 $$\binom{9}{2, \: 3, \: 1, \: 3} = \frac{9!}{2!3!1!3!} = 5040$$
• And for a: $$(2j+k+3l+m)^9$$ you would do: $$\binom{9, \: 2, \: 3}{2, \: 3, \: 1, \: 3} = \frac{9!2!3!}{2!3!1!3!} = 60 480$$ is that correct? – Det Sep 25 '17 at 14:05
• Uhhh.. seems this was accidentally correct? The correct way seems to be to: $$\binom{9}{2, \: 3, \: 1, \: 3}(2j)^2k^33lm^3 = \frac{9!}{2!3!1!3!}(2j)^2k^33lm^3 = ... = 60480j^2k^3lm^3$$ – Det Sep 25 '17 at 14:38
• Yeah, the correct way is the second one, and you're right: the results are equal for a mere coincidence, specifically because $2!3! = 2^2 \cdot 3$... – cip999 Sep 25 '17 at 15:14