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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- Jun 7th 2007, 06:00 AMlkaflI 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? - Jun 7th 2007, 06:30 AMThePerfectHacker
- Jun 7th 2007, 07:18 AMtopsquark
Here's the long way. It also gives slightly more information about the number of possible solutions.

So if

then

for some value of n.

We may immediately cancel the n!, giving:

<-- Divide both sides by 6!

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

Now expand and get everything all to one side:

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

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:

So as ThePerfectHacker said, n = 15 is a solution. By either synthetic or long division, this means that

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 AMSoroban
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

Clear denominators: .

I'll take baby-steps . . .

Divide by

Divide by 6! . . . . . . [ . Remember that!]

We have:. . . . . .

We have: the product of three consecutive integers is 504.

. . Look familiar? . . . They must be: 7, 8, 9.

Therefore: .