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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.

2 Answers 2

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$$(x^2+y^5+z^3)^{30} = ((x^2 + y^5) + z^3)^{30} = \sum^{30}_{k=0} {30\choose k} (x^2 + y^5)^{30-k} z^{3(k)}$$

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.

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    Thank you so much for fast answer – 2017-01-28
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    Welcome. $\Large\ddot \smile $ – 2017-01-28
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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 $$ \left( \begin{matrix} n\\ khl\\ \end{matrix} \right):=\frac{n!}{k!\cdot h!\cdot l!} $$