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Math Help - Discrete and Combinatorial Math - Combination with Repetitions

  1. #1
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    Discrete and Combinatorial Math - Combination with Repetitions

    I need a bit of help with a problem. I am pretty sure I know what formula I should use, I just don't know how.
    Question:
    Find the coefficient of v^2w^4xz in the expansion of (3v + 2w + x + y + z)^8
    I think I need to use (a_1 + a_2 + ... + a_n)^r = \Sigma \binom{r}{x_1 x_2 ... x_n} a^{x_1}_1 a^{x_2}_2 ... a^{x_n}_n but I am not sure. Any help would be great! I have already looked thought the book and there isn't any other problems of this sort that I can find, and this book does not have a students solution manual.
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  2. #2
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    In (x_1+...+x_k)^n the coefficient of x_1^{a_1}...x_k^{a_k} is \frac{n!}{(a_1)!...(a_k)!}.

    This means the coefficent here is: \frac{8!}{2!4!1!1!}

    EDIT: Wrong answer. Also, it should be k=5 and not k=4.
    Last edited by ThePerfectHacker; January 26th 2008 at 07:26 PM.
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  3. #3
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    \frac{8!}{2!4!1!1!} gives me 840. I have derive and when I expand v^2w^4xz I get 120960 for the coeffiction. Are you sure about that solution?
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    In (x_1+...+x_k)^n the coefficient of x_1^{a_1}...x_k^{a_k} is \frac{n!}{(a_1)!...(a_k)!}.

    This means the coefficent here is: \frac{8!}{2!4!1!1!}
    TPH has made an error here (and look, out the window, a flying pig).

    The formula (called the multinomial theorem) is correct, but its application to your specific question needs tweaking: you have 3x_1 rather than x_1 and 2x_2 rather than x_2 .....

    Think about the small changes that will make to the answer given by TPH.
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  5. #5
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    I got it!
    840 \times 3^2 \times 2^4 = 120960

    Thank you guys so much, I finally understand this. Now I just wish you guys were here to help me go over my notes and help me study Oh well. Thank you. I will be coming back when I need help with things, this is a lot better help than I get from other people.
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