## anonymous one year ago solve for n P(n,3) = 60

1. Luigi0210

I'm assuming the P means Permutations.. do you know the formula?

2. anonymous

yes

3. anonymous

P(n,r) = n!/(n-r)!

4. Luigi0210

So I'm assuming they want us to work backwards.. $$\large P(n, r ) = 60 = \frac{n!}{(n-3)!}$$ And we know that that $$n> 3$$

5. anonymous

yeah, for sure :)

6. Luigi0210

I don't think they'd make the problem too difficult.. so I'd suggest just plugging in values for n that are greater than 3. So if we try 4; 4! would equal 24, which is too small and not equal to 60. So 4 won't work. Now trying 5: $$\large P(5, 3) = \frac{5!}{(5-3)!} = \frac{120}{2!}$$ And what's 120/2?

7. anonymous

how to solve this then? i'm required to show a solution

8. anonymous

$0=n ^{3}-3n ^{2}+2n-60$

9. anonymous

i have went all until this equation, and i'm stuck

10. anonymous

but 5's gonna work for that equation for sure

11. Luigi0210

I'm not sure if they want you to go into it that deep, I think just plugging in values would suffice

12. anonymous

yeah they do! and it's HORRIBLE

13. anonymous

but thanks anyway :)