Finding the number equal to coefficient of polynomial I hope I've asked right question for this problem.
I've got a number $$ \frac{(30!)}{(5!)^2*20!}$$
And I got questions about this number:


*

*Is this number equal to the coefficient $x^{10}*y^{100}*z^{15}$ in polynomial $ (x^2+y^5+z^3)^{30} $

*Is this number equal to the coefficient $x^{5}*y^{5}*z^{20}$ in polynomial $ (x+y+z)^{30} $


And the answers are, yes they are equal in this coefficient.
I was searching for this in some of books and I couldn't find the example of doing this. I would be very thankful if somebody could show me how should it be done.
 A: These are special cases of a generalization of Newton's binomial formula:
$$
(a+b+c)^n=\sum_{\stackrel{k+h+l=n}{0\leq k,h,l\leq n}}\left(
\begin{matrix}
        n\\
        khl\\
 \end{matrix}
\right)a^kb^hc^l,
$$
where 
$$
\left(
\begin{matrix}
        n\\
        khl\\
 \end{matrix}
\right):=\frac{n!}{k!\cdot h!\cdot l!}
$$
A: $$(x^2+y^5+z^3)^{30} = ((x^2 + y^5) + z^3)^{30} = \sum^{30}_{k=0} {30\choose k} (x^2 + y^5)^{30-k} z^{3(k)}$$

The (k+1)th term in a binomial expansion of $(x+y)^n$ can be found by, 
$$T_{k+1} = {n \choose k}x^{n-k}y^k$$  

Putting $k = 5$ for the expansion of question, we get 
$$T_{6} = {30 \choose 5}(x^2 + y^5)^{25}z^{3*5}\tag{1}$$  
Now consider the expansion of $(x^2 + y^5)^{25}$
which is, $$\sum^{25}_{k=0} {25\choose k} x^{2(25-k)} + y^{5(k)}$$
Now put $ k = 20$ in the formula. 
we get 
$$T_{21} = {25\choose 20}x^{10}y^{100}\tag{2}$$
From (1) and (2)
Coefficient is $${25\choose 20}{30 \choose 5} = {(25)!\over 5! \times (20)!} \times {(30)!\over 5! \times (25)!} = {(30)!\over (5!)^2\times(20)! }$$
Same can be done for the other question of ours.
