# Homework Help: Permutations Problem

1. Jul 11, 2010

1. The problem statement, all variables and given/known data
Solve for n.

$$(14)_{n}P_{3}=_{n+2}P_{4}$$

2. Relevant equations
$$_{n}P_{r}=\frac{n!}{(n-r)!}$$

3. The attempt at a solution

First I write the problem with the equations written out

$$\frac{14n!}{(n-3)!}=\frac{(n+2)!}{(n-2)!}$$

I'm not quite sure how to isolate the n with all of the factorials going on.
If I subtract the RHS over to the left I get:

$$\frac{(n-2)!14n!-(n+2)!(n-3)!}{(n-2)!(n-3)!}$$

Not sure if this is helpful or not.

Last edited: Jul 11, 2010
2. Jul 11, 2010

### vela

Staff Emeritus
Expand the factorials out a bit and cancel terms. For example,

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

3. Jul 11, 2010

Do you have (n-1)! in the denominator by accident? because it really comes out to be (n-2)!.

$$\frac{14(n-3)!(n-2)(n-1)(n)}{(n-3)!}=\frac{(n-2)!(n-1)(n)(n+1)(n+2)}{(n-2)!}$$

Some canceling and simplifying and you'll get

$$0=\frac{(n+1)(n+2)}{14(n-1)}$$

and that dosen't divide nicely at all. Am I doing something wrong?

4. Jul 11, 2010

### vela

Staff Emeritus
It was just an example; it had nothing to do with the problem.
You don't need to divide. If the fraction is 0, that means the numerator is 0. What values of n make the top equal to 0?

Actually, the LHS shouldn't be 0. Check your work. I think your denominator is off too.

5. Jul 11, 2010

$$14(n-2)(n-1)(n)=(n-1)(n)(n+1)(n+2)$$

then I divided both sides by (n-2)(n-2)(n) to cancel terms and set one side equal to zero so I can solve for n.

6. Jul 11, 2010

### vela

Staff Emeritus
You can't just arbitrarily set one side equal to 0. If you just cancel the factors common to both sides, you get

$$14(n-2) = (n+1)(n+2)$$

How would you solve that?

7. Jul 11, 2010

### vela

Staff Emeritus
Yes, as long as you're not dividing by 0. I added to my previous post, so you may want to look at it again.

8. Jul 11, 2010

$$0=n^2-11n+30$$
$$0=(n-5)(n-6)$$