Coefficient problem using multinomial theorem

Question

i want to solve this: consider $(x+y+z)^n$, let $n=1000$ the coefficient of $x^{320}y^{410}z^{270}$ can be written as $\binom{a}{b} \cdot \binom{c}{d}$.

find $a,b,c,d \in \mathbb{N}$

my attempt is using the multinomial theorem, so I get that the coefficient of $x^{320}y^{410}z^{270}$

$\frac{1000!}{320! \cdot 410! \cdot 270!}=\binom{a}{b} \cdot \binom{c}{d}=\frac{a! \cdot b!}{b!\cdot (a-b)!\cdot d! \cdot (c-d)!}$

But from here i cannot find some relation for $a,b,c$ and $d$.

Answer 1

You have the right idea. As @Henry's question comment indicates, there's more than one answer. From what you started, here's how to get one of those answers:

$$\begin{equation}\begin{aligned} \frac{1000!}{320! \cdot 410! \cdot 270!} & = \left(\frac{1000!}{320! \cdot 410! \cdot 270!}\right)\left(\frac{730!}{730!}\right) \\ & = \left(\frac{1000!}{270!\cdot 730!}\right)\left(\frac{730!}{320!\cdot 410!}\right) \\ & = \binom{1000}{270} \cdot \binom{730}{320} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Answer 2

Well, ${1000\choose 320,410,270} = {1000\choose 320}\cdot {{1000-320}\choose 420}\cdot {{1000-320-410}\choose 270}$, where the last coefficient is $1$.

Answer 3

Hint: Start from your left-hand side and find a representation with binomial coefficients by properly expanding numerator and denominator. We obtain \begin{align*} \frac{1000!}{320! \cdot 410! \cdot 270!}=\frac{1000!}{320!(1000-320)!}\,\frac{(1000-320)!}{410!\cdot 270!} \color{blue}{=\binom{1000}{320}\binom{680}{410}} \end{align*}