Thread: Solving for permutation on one side of the equation and a combination on the other

1. Solving for permutation on one side of the equation and a combination on the other

So far i've been asked to solve for n in either a Permutation or a Combination. I have a question that involves both. Any help with the following equation would be much appreciated (even a way to simplify it?).

Here it is:

Solve for n if P(n,3) = 2C(n,2), n belongs to N

Thanks to anyone who replies.

2. $\displaystyle \frac{n!}{(n-3)!} = 2 \ \frac{n!}{2!(n-2)!}$

$\displaystyle (n-2)! = (n-3)!$

$\displaystyle (n-2)(n-3)! = (n-3)!$

$\displaystyle n-2 = 1$

$\displaystyle n=3$

3. I was just wondering what you divided by in the second line to come to (n-2)! = (n-3)!,

I also have another question and you obviously know what you are doing:

Solve for (n+2)!/n!=56

You are awesome!

4. Originally Posted by jake.davis
I was just wondering what you divided by in the second line to come to (n-2)! = (n-3)!,

Solve for (n+2)!/n!=56
Note that the 2's divide off as well as the n!'s. Then the denominators must be equal.

Now for the next problem note that: $\displaystyle \frac{(n+2)!}{n!}=\frac{(n+2)(n+1)n!}{n!}$

5. Thanks Plato,
For the second question the n!'s would cancel out leaving me with:

(n+2)(n+1)=56