Permutations with Standard Form Equations: Solving for n

[Zinger]

Homework Statement

the problem states solve for n.
nP4 = 8(nP4)

Homework Equations

no relevant equat. i can think of?

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 doesn't seem right so I am quite stuck. help please?

 

[Mentallic]

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.

 

[Zinger]

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.

yeah, this is the stage I am 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)

 

[Mentallic]

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).

 

[Zinger]

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).

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?

 

[Mentallic]

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.

 

[Mentallic]

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

i then tried to isolate the N by adding the 4 onto the other side giving me (n+1)!/8 = n!

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)!

 

[Zinger]

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)!

ooh okay thanks :)
you helped me a lot, there are some aspects i still don't understand however, but i'll ask my teacher. youre a big help.