Math Is Fun Forum

Synthetic.Butterfly

Member

Registered: 2007-09-25

Posts: 6

Permutations and Combinations help...

So, I'm going through my HW and I'm kind of stuck on a problem. I have no idea if I'm doing this right or not.

Q: If P(n, 4) = 360 find n

I got to P(n, 4) = n!/(n-4)! = 360

but I seriously have no Idea where to go from here...  someone help please

ok... edit* second problem...

C (8, 6) = P(8, 6)/6! = 8!/ 6! (8-6)! = 8!/ 6!×2! = 20160/720 = 28

It would help alot if you could explain why each thing is as it is.

Edit* Never mind second problem, I messed up on the equation, I'm fixing it and posting the solution in this edit.

Last edited by Synthetic.Butterfly (2007-10-01 13:31:15)

Synthetic.Butterfly

Member

Registered: 2007-09-25

Posts: 6

Re: Permutations and Combinations help...

come on people.. I'm confused

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Permutations and Combinations help...

Well, note that 7 does not divide 360. So n cannot be 7 or greater – i.e. n ≤ 6. Try [sup]6[/sup]P[sub]4[/sub].

It worked! Therefore n = 6.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Permutations and Combinations help...

Write out the prime factorization of 360:

360 = 5*2*36 = 2^3*3^2*5

Now we have n! / (n-4)! So what we want is n*(n-1)*(n-2)*(n-3).  We have 6 primes to move around, so at least 2 of these numbers must be single primes, and the other two can be at most 3 primes.  By simple inspection, we can remove the three primes from our list of options.  This is because the smallest number of three primes is 2^3 = 8, and even that's too big.  Obviously anything larger isn't going to help.

It should be immediate that 5 is by itself.  So we have to have either 4 or 6 as well.  If we have 6, then what's left is 2^2 and 3, which make 3, 4, 5, and 6.  If we went the other way, we reach the same conclusion just as quickly.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Synthetic.Butterfly

Member

Registered: 2007-09-25

Posts: 6

Re: Permutations and Combinations help...

Ok, I understand where I went wrong... I kept forgetting to take in account that it is (n-4)! meaning that I was missing a step while computing it. Gosh, don't I fell smart.