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Math Help - I got stuck on this problem: C(n,6)=C(n,9)

  1. #1
    lkafl
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    I got stuck on this problem: C(n,6)=C(n,9)

    i know how to set it up and all but when I get to a certain point... I got stuck..

    Help please?
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  2. #2
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    Quote Originally Posted by lkafl View Post
    i know how to set it up and all but when I get to a certain point... I got stuck..

    Help please?
    The theorem says that,
    {n\choose m} = {n\choose n-m}

    You have,
    {n \choose 9} = {n\choose 6}
    It seems that,
    n-9 = 6 \Rightarrow n=15
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lkafl View Post
    i know how to set it up and all but when I get to a certain point... I got stuck..

    Help please?
    Here's the long way. It also gives slightly more information about the number of possible solutions.

     _nC_6 = \frac{n!}{6!(n - 6)!}
     _nC_9 = \frac{n!}{9!(n - 9)!}

    So if
    _nC_6 = \, _nC_9
    then
    \frac{n!}{6!(n - 6)!} = \frac{n!}{9!(n - 9)!}
    for some value of n.

    We may immediately cancel the n!, giving:
    \frac{1}{6!(n - 6)!} = \frac{1}{9!(n - 9)!}

    9!(n - 9)! = 6!(n - 6)! <-- Divide both sides by 6!

    \frac{9!}{6!} (n - 9)! = (n - 6)!

    \frac{9 \cdot 8 \cdot 7 \cdot 6!}{6!} (n - 9)! = (n - 6)!

    9 \cdot 8 \cdot 7 (n - 9)! = (n - 6)!

    504(n - 9)! = (n - 6)! <-- Divide both sides by (n - 9)!

    504 = \frac{(n - 6)!}{(n - 9)!}

    504 = \frac{(n - 6)(n - 7)(n - 8)(n - 9)!}{(n - 9)!}

    504 = (n - 6)(n - 7)(n - 8)

    Now expand and get everything all to one side:
    n^3 - 21n^2 + 146n - 840 = 0

    Our advantage (and only advantage) here in solving this is that we know n must be a positive integer and, to make any sense, n \geq 9.

    So let's simply try different values of n one by one until we find a solution. (If you wish to narrow the list a bit, the rational roots theorem says that n must be a factor of 840 as well.) So here we go:
    9^3 - 21 \cdot 9^2 + 146 \cdot 9 - 840 = -498
    10^3 - 21 \cdot 10^2 + 146 \cdot 10 - 840 = -480
    11^3 - 21 \cdot 11^2 + 146 \cdot 11 - 840 = -444
    12^3 - 21 \cdot 12^2 + 146 \cdot 12 - 840 = -384
    13^3 - 21 \cdot 13^2 + 146 \cdot 13 - 840 = -294
    14^3 - 21 \cdot 14^2 + 146 \cdot 14 - 840 = -168
    15^3 - 21 \cdot 15^2 + 146 \cdot 15 - 840 = 0

    So as ThePerfectHacker said, n = 15 is a solution. By either synthetic or long division, this means that
    n^3 - 21n^2 + 146n - 840 = (n - 15)(n^2 - 6n + 56) = 0

    As it happens that last factor gives only complex values of n, so n = 15 is the only solution.

    -Dan
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  4. #4
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    Hello, lkafl!

    i know how to set it up and all
    but when I get to a certain point, I got stuck. .I did, too

    C(n,6) \:=\:C(n,9)
    We have: . \frac{n!}{6!(n-6)!} \;=\;\frac{n!}{9!(n-9)!}

    Clear denominators: . 9!n!(n-9)! \;=\;6!n!(n-6)!


    I'll take baby-steps . . .

    Divide by n!\qquad\qquad 9!(n-9)! \;=\;6!(n-6)!

    Divide by 6! . . . 9\!\cdot\!8\!\cdot\!7(n-9)! \;=\;(n-6)! . . . [ 9\!\cdot\!8\!\cdot\!7 = 504 . Remember that!]

    We have:. . . . . . 504(n-9)! \;=\;(n-6)(n-7)(n-8)\!\cdot\!(n-9)!

    \text{Divide by }(n-9)!\qquad\qquad\;\;\; 504 \:=\:\underbrace{(n-6)(n-7)(n-8)}_{\text{3 consecutive integers}}


    We have: the product of three consecutive integers is 504.
    . . Look familiar? . . . They must be: 7, 8, 9.

    Therefore: . n = 15

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