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Math Help - combinations

  1. #1
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    combinations

    [note...n choose k entails the following: n!/(k!(n-k)!)

    i have to simplify following statement but don't know how to go about it:

    (n choose 4) + 2(n choose 5) + (n choose 6)

    so if anyone could help me, that would be great
    Last edited by buckaroobill; December 16th 2006 at 02:20 PM.
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  2. #2
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    Quote Originally Posted by buckaroobill View Post
    [note...n choose k entails the following: n!/(k!(n-k)!)

    i have to simplify following statement but don't know how to go about it:

    (n choose 4) + 2(n choose 5) + (n choose 6)

    so if anyone could help me, that would be great
    For sufficently large n,

    \frac{n!}{4!(n-4)!}+\frac{2n!}{5!(n-5)!}+\frac{n!}{6!(n-6)!}
    Common denominator is,
    6!(n-4)!
    Thus,
    \frac{6\cdot 5 \cdot n!}{6!(n-4)!}+\frac{2\cdot 6\cdot n!(n-4)}{6!(n-4)!}+\frac{n!(n-5)(n-4)}{6!(n-4)!}
    Now you can combine fractions
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  3. #3
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    Hello, buckaroobill!

    My approach is slightly different . . .


    Simplify: .  \binom{n}{4} + 2\binom{n}{5} + \binom{n}{6}

    We have: . \frac{n!}{4!(n-4)!} + 2\frac{n!}{5!(n-5)!} + \frac{n!}{6!(n-6)!}

    . . = \;\frac{n(n-1)(n-2)(n-3)}{4!} + \frac{2n(n-1)(n-2)(n-3)(n-4)}{5!}  + \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}

    Factor: . n(n-1)(n-2)(n-3)\left[\frac{1}{4!} + \frac{2(n-4)}{5!} + \frac{(n-4)(n-5)}{6!}\right]

    Common denominator: . n(n-1)(n-2)(n-3)\left[\frac{1}{4!}\!\cdot\!\frac{5\!\cdot\!6}{5\!\cdot\!  6} + \frac{2(n-4)}{5!}\!\cdot\!\frac{6}{6} + \frac{(n-4)(n-5)}{6!}\right]

    . . = \;\frac{n(n-1)(n-2)(n-3)}{6!}\left[30 + 12(n-4) + (n-4)(n-5)\right]

    . . = \;\frac{n(n-1)(n-2)(n-3)}{6!}\left(n^2 + 3n + 2\right)  \;=\;\frac{n(n-1)(n-2)(n-3)}{6!}(n+1)(n+2)

    . . = \;\frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{6!} \;=\;\boxed{\binom{n+2}{6}}

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