Hey Guys, been a while since I have done a problem as such.

I have a lock with three selections of numbers, each selection goes 1-30 (giving 30 possible numbers). How many different combinations are possible?

I used the general A^{n}where n is the selections (3) and A is the possible numbers (30) giving me

A^{n }= 30^{3}=27,000.

The next part was tricky, I need to find how many possible numbers when no two numbers can be repeated by any combination.

I think this is where I use

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

which gave me

30!/(30-3)! = 24360.

Does this look right to you guys?

Thanks for your help!