Both parts are correct.
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 An where n is the selections (3) and A is the possible numbers (30) giving me
An = 303=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!
I can use this same principle with say letters too, e.g. THEORY arranged in any order would be
6! = 720.
Then how about it I needed T and H to stay together, but could switch order, e.g. TH and HT are ok. Could I just say...
1.TH or HT
Then say 5! = 120, but as TH can also be HT, double the result to give me the answer? so 240 possible orders.