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Thread: special case of combinatorics

  1. #1
    Aug 2011

    special case of combinatorics

    number of ways in which it is possible to make a selection from m+n+p=N things,where p are alike of one kind, m alike of another kind and n alike of another kind,taken r at a time is given by co-efficient of x^r in the expansion of
    (1+x+x^2+x^3+......+x^m)(1+x+x^2+.....x^n)(1+x+x^2 +x^3+.......x^p).
    How is this established?Explain me the Q.E.D.
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  2. #2
    MHF Contributor
    Oct 2009

    Re: special case of combinatorics

    A selection of $\displaystyle r$ objects is determined by the number of objects of each kind. So, the number of such selections is the number of triples $\displaystyle (i,j,k)$ such that $\displaystyle i+j+k=r$ and $\displaystyle i\le i\le m$, $\displaystyle 1\le j\le n$ and $\displaystyle 1\le k\le p$. When you look at the polynomial, this is exactly the coefficient of $\displaystyle x^r$. Indeed, to get $\displaystyle x^r$ one needs to choose some $\displaystyle x^i$ from the first factor, $\displaystyle x^j$ from the second factor and $\displaystyle x^k$ from the third factor so that $\displaystyle i+j+k=r$. The number of such choices equals the coefficient of $\displaystyle x^r$.
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