Question 1093789
<br>If the number is divisible by 5, then the last digit is either 0 or 5.  So you have two cases to analyze.<br>
If the last digit is 0, then there are 9 choices for one of the other digits and then 8 choices for the third digit.  By the fundamental counting principle, the number of numbers less than 1000, divisible by 5, ending in 0, and containing 3 different digits is 9*8 = 72.<br>
If the last digit is 5, there are still 9 choices for one of the other digits.  That is because the problem said "numbers less than 1000", not "3-digit numbers" -- so the first digit can be 0.  And then again there are 8 choices for the last digit, making again 9*8=72 numbers less than 1000, divisible by 5, ending in 5, and containing 3 different digits.<br>
So there are a total of 72+72=144 numbers that are less than 1000, are divisible by 5, and contain 3 different digits.