Find the number of three-digit numbers which are divisible by 5 and have distinct digits

Explanation

A number is divisible by 5 if digit at unit place is either '0' or '5'.
So, we have following two cases :
Case I : Digit '0' at unit place

In this case hundred's place, can be filled in 9 ways and tens place can be filled in 8 ways.
So, number of numbers `=9xx8=72` ways. Case II : Digit '5' at unit place

In this case hundred place can be filled in 8 ways (as zero cannot be used).
Tens place can be filled in 8 ways (as zero can be used)
So, number of number `=8xx8=64` ways.
Therefore, total number of numbers =72+64=136