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

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