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

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.

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HallsofIvy
Homework Helper
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!

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?

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HallsofIvy
Homework Helper
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.

I trie with Latex:

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

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

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

What do you mean by "and then cancel"???