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

• Jun 7th 2007, 06:00 AM
lkafl
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..

• Jun 7th 2007, 06:30 AM
ThePerfectHacker
Quote:

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

The theorem says that,
${n\choose m} = {n\choose n-m}$

You have,
${n \choose 9} = {n\choose 6}$
It seems that,
$n-9 = 6 \Rightarrow n=15$
• Jun 7th 2007, 07:18 AM
topsquark
Quote:

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

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
• Jun 7th 2007, 07:22 AM
Soroban
Hello, lkafl!

Quote:

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: . $n = 15$