Assumption: The 5-digit number will not begin with 0.
The 5 digit number can be formed in the following number of ways:
5x5x4x3x2
The first(left most) digit can be either 1,2,3,4 or 5.
The 2nd digit can be either of of the remaining 5 digits.
The 3rd digit can be either of of the remaining 4 digits.
etc… (this is the peagon hole concept for counting)
A number divisible by 5 can be formed with the last digit ending with either 5 or 0.
Case 1) Number ending with digit 0:
If the last digit is 0, then the first 4 digits can be arranged in the following number of ways:
5x4x3x2
The first(left most) digit can be either 1,2,3,4 or 5.
The 2nd digit can be either of of the remaining 4 digits.
The 3rd digit can be either of of the remaining 3 digits.
The 4th digit can be either of of the remaining 2 digits.
Case 2)Number ending with digit 5:
If the last digit is 5, then the first 4 digits can be arranged in the following number of ways:
4x4x3x2
The first(left most) digit can be either 1,2,3, or 4 (0 is not allowed, and 5 is already in use)
The 2nd digit can be either of of the remaining 4 digits.
The 3rd digit can be either of of the remaining 3 digits.
The 4th digit can be either of of the remaining 2 digits.
Therefore, the probability of having a 5-digit number divisible by 5 is:
{(5x4x3x2) + (4x4x3x2)} / (5x5x4x3x2)
=(5 + 4) / (5x5)
=9/25
Choose E.
-BM-
