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