Solving for n in Permutation Equations

[themadhatter1]

Homework Statement

Solve for n.

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

Homework Equations

$$_{n}P_{r}=\frac{n!}{(n-r)!}$$

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.

 

[vela]

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

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

 

[themadhatter1]

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

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

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

Ok, I see your method,

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

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?

 

[vela]

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

It was just an example; it had nothing to do with the problem.

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

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

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.

 

[themadhatter1]

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.

I originally had

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

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.

Your right about the denomonator though it should be (n-2)14

 

[vela]

I originally had

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

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.

Your right about the denomonator though it should be (n-2)14

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?

 

[vela]

well I didn't just set a side equal to zero. I actually divided both sides of the equation by the same thing. Thats ok right?

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.

 

[themadhatter1]

Ahh Foil out and move to one side.

so you get

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

therefore n=5,6 and that's what the answer is supposed to be.

Thanks!