Thread: 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?

Thanks for your help!

Both parts are correct.

Thanks!

You could use the fundamental counting rule:

a) 30·30·30 = 27000

b) 30·29·28 = 24360

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?

Using the fundamental counting rule, how many choices do you have for the first seat, second seat, and so on?

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!

Yes 6! is correct. There are n! ways to order n objects.

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.

Yes, that's correct.

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