Need math expert..

[GoldenBear]

Not so much an expert, but someone =-)

I'm having trouble solving this..

P is the probability thingie..

7 [P(n,5)] = P(n,3) * P(9,3)

So now I have..

7(n! / (n-5)!) = 504 (n! / (n-3)!)

I'm stuck now =-(

 

[bolomite]

Expand those factorials: (n!) = (n)(n-1)(n-2) ...etc., do this for all factorial terms and cancel what you can. Looks like you'll then wind up with some sort of polynomial in n, maybe 4th degree.

 

[Cerebus451]

n!/(n-5)! = n * (n - 1) * (n - 2) * (n - 3) * (n - 4)

likewise:
N!/(n-3)! = n * (n - 1) * (n - 2)

so....

7 * (first line) = 504 * (second line)
divide both sides by n * (n - 1) * (n - 2)

7 * (n - 3) * (n - 4) = 504
(n - 3) * (n - 4) = 72
n^2 - 7n + 12 = 72
n^2 - 7n - 60 = 0

take it away from there.
I think. Too late at night for me to double check my work.
Don't forget n cannot be 0, 1, or 2 otherwise that whole division thing is shot.
I believe n = 12 is one solution.

 

[bolomite]

Ah, I spoke before I thought it through. I believe you'll end up with just a quadratic after simplification. I get n = -5 and 12.

 

[Cerebus451]

You go the other one bolomite. I was working it through in my head and was only able to come up with the 12 answer without having to write anything down.