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Math Help - A binom question

  1. #1
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    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.
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  2. #2
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    Re: A binom question

    Quote Originally Posted by sunrise View Post
    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
    Last edited by romsek; December 14th 2013 at 11:11 PM.
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  3. #3
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    Re: A binom question

    Yes romsek,

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

    I prefer, waiting of friends explanations. Thank you.
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  4. #4
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    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)!}
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  5. #5
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    Re: A binom question

    Thank you very much
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