# Thread: solving a combination equation

1. ## solving a combination equation

I would like some help with a question that I do not fully understand.

The question is to solve the following for n:

4•C(n,2) = C(n+2, 3)

I know that some form of the equation:

n!
_______

r!(n-r)!

but i can't seem to solve it. I also know that the answer is 2 or 7.

thanks, john.

2. Originally Posted by drew487
I would like some help with a question that I do not fully understand.

The question is to solve the following for n:

4•C(n,2) = C(n+2, 3)

I know that some form of the equation:

n!
_______

r!(n-r)!

but i can't seem to solve it. I also know that the answer is 2 or 7.

thanks, john.
$4 \cdot \frac{n!}{2!(n-2)!} = \frac{(n+2)!}{3!(n-1)!}$

$4 \cdot \frac{n!}{2!(n-2)!} = \frac{(n+2)(n+1)n!}{3 \cdot 2!(n-1)(n-2)!}$

divide out the common factors ...

$4 = \frac{(n+2)(n+1)}{3(n-1)}$

can you finish?

3. yess thank you, you cleared that up for me.. I was it a much more complicated and well.. dumb way. But i get it now!