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Math Help - special case of combinatorics

  1. #1
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    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
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    Re: special case of combinatorics

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