Wait! I was wrong. See this link:https://web2.0calc.com/questions/help-please_56105
+118609 Guest 1: How did you get this answer
Guest 2: Why do you think the other answer was wrong.
Hi Melody: I don't know how Guest #1 got that answer which is CORRECT.
You arrive at the answer as follows:
1- Exclude all 2's and 3's from 2nd position to the left which leaves you with 8 numbers(0, 1, 4,5, 6, 7, 8, 9).
2 - This exclusion applies to the 3rd, 4th and 5th position.
3 - Therefore you have: 8 x 8 x 8 x 8 =4096
4 - Since there are only 7 numbers {1, 4, 5, 6, 7, 8, 9} to contend with, because we are excluding all 20,000 and 30,000 numbers, then we have: 7 x 4096 =28,672 numbers that must be excluded from the total of 90,000 5-digit integers.
5 - 90,000 - 28,672 = 61,328 numbers that must have either a 2 or a 3 or both in them.
6 - Just to confirm this, I wrote a short computer code to count them and got exactly this number of 61,328.
