Integers 0-9999999 containing 2 3's and 2 5's

[SneakyArab]

Homework Statement

How many integers from 0-9,999,999 have exactly two 3's and two 5's as digits.

Homework Equations

I'm not really sure...

The Attempt at a Solution

The answer is 107520, if I'm not mistaken. I made a program to count it up for me, so I'm fairly sure that that is the correct answer. I'm just trying to figure out how to do it manually now.

So I know there are 7 digits, 4 of which are 3, 3, 5, 5, and the other 3 are 0,1,2,4,6,7,8,9.
So the remaining 3 digits each have 8 possibilities. So there are 8^3 different combos for the other 3 digits. But then all of the digits can be rearranged, so I thought it would be 7! * 8^3. That was not even close to correct.

 

[AUMathTutor]

Well, each of the 7 digits can be any digit from 0...9. So we have the numbers

0 0 0 0 0 0 0
0 0 0 0 0 0 1
...
9 9 9 9 9 9 8
9 9 9 9 9 9 9

Clearly, there are 10^7, or 10 million, of these.

So, you have to pick 2 from 7 to call "3". Then you have to pick 2 from the remaining 5 places to call "5". Then you can choose any of 8 digits to fill the last three spots.

If you use this information, you WILL get the right answer. If I were to do the next step, you would get the answer. That's how close you are now. Focus on the logic I used and understand that, as that's what's most important.

 

[Dick]

Not all rearrangements give rise to different numbers. You are over counting. First pick two positions out of the 7 to place the 3's. How many ways to do that?

 

[SneakyArab]

Ooooh ok, I see.
so its 7 choose 2 * 5 choose 2 * 8^3
Thank you so much!