My question is:
Consider the set of 4 digit numbers 1000,1001.....9999. How many of these numbers has exactly three zeros behind it,
Just these 9
{1000,2000,3000,4000,5000,6000,7000,8000,9000}
and how many of the numbers has distinct digits i.e. No numbers are repeated.
Choose the first digit any of 9 ways (it can't be 0.
Choose the second digit any of 9 ways, (it can be 0, but it can't be the one you selected for the first digit)
Choose the third digit any of 8 ways, (it can't be either of the ones you selected for the first two digits)
Choose the fourth digit any of 7 ways, (it can't be either of the ones you selected for the first three digits)
That's 9·9·8·7 = 4536
Edwin
