# Standard form permutation

1. Feb 18, 2010

### Zinger

1. The problem statement, all variables and given/known data
the problem states solve for n.
nP4 = 8(nP4)

2. Relevant equations

no relevant equat. i can think of?

3. The attempt at a solution
my attempt at this was nP4 = 8(nP3)
idk what i tried to do, but i tried to get it in standard form i guess: n!/ (n-4) = 8(n!)/(n-3)
it doesnt seem right so im quite stuck. help please?

2. Feb 18, 2010

### Mentallic

Re: permutations

No, you are on the right track. However you forgot the extra factorial signs:

$$\frac{n!}{(n-4)!}=\frac{8n!}{(n-3)!}$$

Now solve for n by using what you know about factorials to simplify the equation.

3. Feb 18, 2010

### Zinger

Re: permutations

yeah, this is the stage im talking about, exactly what am i suppose to do to simplify.
im thinking to cross mult. b/c its a proportion and turn it into a quad form? the problem is, the factorials; how can I simplify with all of them? and there are no factorials for N! (that I can solve out)

4. Feb 18, 2010

### Mentallic

Re: permutations

Use the fact that $n!=n(n-1)!=n(n-1)(n-2)!$ etc. and solve the equation like you would any other equation (dividing by n! on both sides would be a good start).

Last edited: Feb 18, 2010
5. Feb 18, 2010

### Zinger

Re: permutations

hmm okay i understand.
i tried that and this was my process/answer:
i cross mult - n!(n-3)!=8n!(n-4)
i then divided (as you suggested) n!(n-3)!/8n! = 8n!(n-4)!/8n!
i then tried to isolate the N by adding the 4 onto the other side giving me (n+1)!/8 = n!

I tried solving like a normal equation by dividing but the N on both sides confuses me, if I do something with it; won't it cancel?
what step did I get wrong?

6. Feb 18, 2010

### Mentallic

Re: permutations

You're making things more complicated than they need to be. Let's start again.

$$\frac{n!}{(n-4)!}=\frac{8n!}{(n-3)!}$$

now, using the rule I told you, we will apply it in a way to make both sides more equivalent in a sense.
Notice that if $n!=n(n-1)!$ then similarly, $(n-3)!=(n-3)(n-4)!$

So we have $$\frac{n!}{(n-4)!}=\frac{8n!}{(n-3)(n-4)!}$$

Now divide both sides by $$\frac{n!}{(n-4)!}$$ and you're cleared of all factorials.

7. Feb 18, 2010

### Mentallic

Re: permutations

Oh and by the way, you went wrong on this line:

From the line before: $$\frac{n!(n-3)!}{8n!} = \frac{8n!(n-4)!}{8n!}$$

Simplified: $$\frac{(n-3)!}{8}=(n-4)!$$

And now using the rule (properly) you should have $$\frac{(n-3)(n-4)!}{8}=(n-4)!$$

And now you can divide through by (n-4)!

8. Feb 21, 2010

### Zinger

Re: permutations

ooh okay thanks :)
you helped me alot, there are some aspects i still dont understand however, but i'll ask my teacher. youre a big help.