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
Hello, lkafl!
We have: .$\displaystyle \frac{n!}{6!(n-6)!} \;=\;\frac{n!}{9!(n-9)!} $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) $
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: .$\displaystyle n = 15$