Permutations Find the values ex. P(n,2)=90

[ClosedCastle]

How can I solve this?

Find the value of n.1. P(n,2)=90 ; 2. P(n,3)=3P(n,2)
I tried like this but I am stuck

1. n! / (n-2)! = 90 and then I don't know ...

2. n!/(n-3!) = 3 n!/(n-2)! ⇔ n! (n-2)! / (n-3)! n! = 3 ⇔ (n-2)! / (n-3)! = 3 here the same As you can see I'm really stuck with this.

 

[HallsofIvy]

You do understand that n!= n(n-1)(n-2)...(3)(2)(1), right?
So (n-2)!= (n-2)...(3)(2)(1).
\frac{n!}{(n-2)!}= \frac{n(n-1)(n-2)...(3)(2)(1)}{(n-2)...(3)(2)(1)}= 90

\frac{(n-2)!}{(n-3)!}= \frac{(n-2)(n-3)...(3)(2)(1)}{(n-3)... (3)(2)(1)}= 3

Cancel!

 

[ClosedCastle]

Ok thanks, I think I understand what you mean,

So 1) n(n-1)=90 then n=10
2) n-2 = 3 then n=5

but if I trie to solve this one it doesn't at up:

3) 2P(n,2)+50=P(2n,2)

2((n!/(n-2))!+50)=2n!/(2n-2)!

Is this correct?

 

[HallsofIvy]

Ok thanks, I think I understand what you mean,

So 1) n(n-1)=90 then n=10
2) n-2 = 3 then n=5

but if I trie to solve this one it doesn't at up:

3) 2P(n,2)+50=P(2n,2)

2((n!/(n-2))!+50)=2n!/(2n-2)!

Is this correct?

Not the way you have written it because you have the parentheses wrong. The last "!" should not be the whole "n!/(n-2)" and 2 does not multiply 50. On the right, it is clearer if you write (2n)! rather than 2n! which can be interpreted as 2(n!). What you should have is
2(n!/(n-2)!+ 50= (2n!)/(2n-2)! or
2\frac{n!}{(n-2)!}+ 50= \frac{(2n)!}{(2n-2)!}

Now write those factorials out in detail and cancel.

 

[ClosedCastle]

I trie with Latex:

2\frac{n!}{(n-2)! + 50 = \frac{2n!}{(2n-2)!}[itex\]

 

[ClosedCastle]

ok, that didn't go the way I expected,hehe

thank you HallsofIvy, now I have to learn how to use LaTeX..

 

[ClosedCastle]

What do you mean by "and then cancel"?