Multinomial theorem problem

Question

I am supposed to determine the coefficient of $$x^{2}z^{3}$$ of expression:$$(x+y+z)^{5}$$. What I did: $$\sum _{i=0}^{5}\binom{5}{i}x^{5-i}\left ( y+z \right )^{i}=\sum _{i=0}^{5}\binom{5}{i}x^{5-i}\sum _{j=0}^{i}\binom{i}{j}y^{i-j}z^{j}$$ and that should be equal to$$\binom{5}{i,j,5-i-j}x^{i}y^{j}z^{5-i-j}$$ Can anyone please explain the last step?

Answer 1

$$\require{cancel} \dbinom{5}{i}\dbinom{i}{j} = \dfrac{5!}{(5-i)!\cancel{i!}}\cdot \dfrac{\cancel{i!}}{j!(i-j)!} = \dfrac{5!}{(5-i)!j!(i-j)!}$$

Now, we re-index:

$$i' = i-j, j'=j$$

This implies $i = i'+j'$, so the RHS of the equation above becomes:

$$\dfrac{5!}{(5-i'-j')!j'!i'!} = \dbinom{5}{i',j',5-i'-j'}$$