# Combination Lock Problem

• Nov 15th 2012, 01:47 PM
Combination Lock Problem
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?

• Nov 15th 2012, 02:02 PM
Plato
Re: Combination Lock Problem
Both parts are correct.
• Nov 15th 2012, 03:04 PM
Re: Combination Lock Problem
Thanks!
• Nov 15th 2012, 03:15 PM
MarkFL
Re: Combination Lock Problem
You could use the fundamental counting rule:

a) 30·30·30 = 27000

b) 30·29·28 = 24360
• Nov 15th 2012, 03:17 PM
Re: Combination Lock Problem
Ah yea I saw that.

Had another question, similar to the one above.
I have 6 people, how many different ways can I seat the 6 people in six seats?

Would that just be a 26​=32?

Or is this a 6! = 720?
• Nov 15th 2012, 03:19 PM
MarkFL
Re: Combination Lock Problem
Using the fundamental counting rule, how many choices do you have for the first seat, second seat, and so on?
• Nov 15th 2012, 03:20 PM
Re: Combination Lock Problem
ah ok got it. I thought it would be 6!, but it just seemed too big for six numbers. Never really thought about it I guess!
• Nov 15th 2012, 03:21 PM
MarkFL
Re: Combination Lock Problem
Yes 6! is correct. There are n! ways to order n objects.
• Nov 15th 2012, 03:29 PM
Re: Combination Lock Problem
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
2.E
3.O
4.R
5.Y

Then say 5! = 120, but as TH can also be HT, double the result to give me the answer? so 240 possible orders.
• Nov 15th 2012, 03:42 PM
MarkFL
Re: Combination Lock Problem
Yes, that's correct.
• Nov 15th 2012, 03:43 PM