# A binom question

• December 14th 2013, 11:57 PM
sunrise
A binom question
Hello,

In the expansion of (x1 + x2 + x3)^10, what is the coefficient of the x^22?

How we can solve this kind of problems can you explain me?

Thank you.
• December 15th 2013, 12:03 AM
romsek
Re: A binom question
Quote:

Originally Posted by sunrise
Hello,

In the expansion of (x1 + x2 + x3)^10, what is the coefficient of the x^22?

How we can solve this kind of problems can you explain me?

Thank you.

just to be clear do you mean

$(x + x^2 + x^3)^{10}$

what you want is the multinomial theorem Multinomial theorem - Wikipedia, the free encyclopedia

and you'd let $x_1=x^1$, $x_2=x^2$, $x_3=x^3$
• December 15th 2013, 12:21 AM
sunrise
Re: A binom question
Yes romsek,

I mean(x^1+x^2+x^3)^10

I prefer, waiting of friends explanations. Thank you.
• January 3rd 2014, 12:16 PM
HallsofIvy
Re: A binom question
I think I would be inclined to do it this way:
$(x^3+ x^2+ x)^{10}= (x^3+ (x^2+ x))^{10}= \sum_{n= 0}^{10} \begin{pmatrix}10 \\ n\end{pmatrix}(x^2+ x)^nx^{30- 3n}$
So that, for every n, we will have $x^{22}$ provided the power of x in $(x^2+ x)^n$ is 30- 3n- 22= 8- 3n. Of course, $(x^2+ x)^n= \sum_{x= 0}^n\begin{pmatrix}n \\ i\end{pmatrix} (x^2)^ix^{n- i}$ so x will have power 8- 3n when 2i+ n- i= i+ n= 8- 3n or i= 8- 4n. The coefficient of that power will be $\begin{pmatrix}n \\ 8- 4n\end{pmatrix}= \frac{n!}{(8- 4n)!(5n- 8)!}$.

Since that is for every n, the coefficient of $x^{22}$ will be $\sum_{n= 0}^{10} \begin{pmatrix}10 \\ n\end{pmatrix}\begin{pmatrix}n \\ 8- 4n\end{pmatrix}= \sum_{n=0}^{10}\frac{10!}{n!(10- n)!}\frac{n!}{(8- 4n)!(5n- 8)!}= \sum_{n=0}^{10}\frac{10!}{(10-n)!(8- 4n)!(5n- 8)!}$
• January 4th 2014, 02:15 AM
sunrise
Re: A binom question
Thank you very much