In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?
please explain
288
please explain
288
If we are looking at odd numbers then the last digit of that number would end in 1 3 5 7 9
1000 is an even number so we don't need to consider it. Lets consider numbers till 999.
For the last digit we can have 1 3 5 7 9 but 5 is not allowed so we have 1 3 7 9= 4 options
For the middle digit we can 0 1 2 3 4 6 7 8 9. All the digits except 5 which equals 9 options
For the first digit we can have 1 2 3 4 6 7 8 9. All except 5 and zero. We cannot include zero as that would make the number a two digit number. so 8 options
So we have 4*9*8=288
