Given the set of digits{2,3,4,7,8,9} how many three-digit numbers can be formed?

Hope there is a magic Formula, that someone knows, otherwise:
You could start with 234, 237, 238, 239. 243, 247, 248, 249, 273, 274,278, 279, 283, 284, 287, 289, 293, 294, 297, and 298.
See if I missed some of the set that starts with a 2. I would think ever how many numbers is found with 2's should be able to be multiplied by 6, as each number should have the same to a starting number.
Couple of good Math Guys/Girls? Could give good input.

If repetition is allowed, then there are

6^{3}=216 numbers that can be formed.

If repetition is not allowed, then there are

6\cdot 5\cdot 4=120 numbers that can be formed.

When forming 3 digit numbers, order is important.

So, P(6,3)=120

Stratmando, you're on the right track. You have 20 listed that start with 2.

There are 6 digits, so 20*6=120

Galactus, I was going to mention you, but didn't wan't to leave someone out so I mentioned no one
Hey, Cool that it works out(20 combinations X 6 = 120).
I love Math and figuring, barely made it through High School. I Love solving problems.

Galactus,
I have a couple of Locks, I lost the Keys, and when pulling the pins, I accidentally dumped them, there are 4 different pins, 2 are the same length, What are the total number of keys to be made to insure 1 of them works?

Thanks, you are ALWAYS good on this stuff. Take care.

Given the set of digits{2,3,4,7,8,9} how many three-digit odd numbers can be formed

Half as many as in your previous question.