this is the question, I'm just not sure how to do it, I'm confused.a 3-digit number can be randomly generated. How many 3 digit numbers are possible if the first number is a 9 and its an even number? And suggestions or tips?:confused: a nine and its even number? Huh?

That does sound a bit bizarre. I think they mean like 934 or 972, which are even numbers that start with 9. I've never seen one like that, so let's think this through.

Don't think of it as 900 and something, and therefore how many of those there are. Think of it like three slots you could place a number in. I always think of the "how many ways" one as being slots that I have to fill in.

If this 3-digit number could be anything, then you would have 10 choices (0-9) for the first slot, the same 10 choices for the second slot, and the same 10 choices for the third slot. Using your multiplication rule, that would be 10x10x10, right?

Well, instead, you already have the number for the first slot. So there's only one choice (outcome) for that one. So that's a 1.

Now, what digits could go into the 2nd slot? Is there a rule about what that 10's column must be? No. So how many outcomes (possibilities) for that one?

Then the third slot has to be even to make the entire 3-digit number even. So how many choices for that one?

I'm fairly sure I'm looking at this correctly. I'm sure someone will correct me if I've gone off base.:p