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Thread: 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,
    $\displaystyle {n\choose m} = {n\choose n-m}$

    You have,
    $\displaystyle {n \choose 9} = {n\choose 6}$
    It seems that,
    $\displaystyle n-9 = 6 \Rightarrow n=15$
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  3. #3
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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.

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

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

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

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

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

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

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

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

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

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

    $\displaystyle 504 = (n - 6)(n - 7)(n - 8)$

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

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

    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:
    $\displaystyle 9^3 - 21 \cdot 9^2 + 146 \cdot 9 - 840 = -498$
    $\displaystyle 10^3 - 21 \cdot 10^2 + 146 \cdot 10 - 840 = -480$
    $\displaystyle 11^3 - 21 \cdot 11^2 + 146 \cdot 11 - 840 = -444$
    $\displaystyle 12^3 - 21 \cdot 12^2 + 146 \cdot 12 - 840 = -384$
    $\displaystyle 13^3 - 21 \cdot 13^2 + 146 \cdot 13 - 840 = -294$
    $\displaystyle 14^3 - 21 \cdot 14^2 + 146 \cdot 14 - 840 = -168$
    $\displaystyle 15^3 - 21 \cdot 15^2 + 146 \cdot 15 - 840 = 0$

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

    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

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

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


    I'll take baby-steps . . .

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

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

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

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


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

    Therefore: .$\displaystyle n = 15$

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