Question 1097086
<br>Since the first digit can't be 0, the number of different 5-digit numbers we can make with these digits is 4*4*3*2*1 = 96.<br>
With these digits, if the number is to be divisible by 4, the possibilities for the sequence of final two digits are
04, 12, 24, or 32<br>
We look at separate cases for each of the possible first digits (1, 2, 3, or 4)
For each case, we determine which of the four possible sequences of final two digits are possible.
For each of the cases we find with a given first digit and allowable last two digits, there are two different ways to place the remaining two digits in positions 2 and 3.<br>
first digit 1: 3 of the 4 sequences of final two digits are possible, so 3*2=6 solutions
first digit 2: only 1 of the 4 sequences of final two digits is possible, so 1*2 = 2 solutions.
first digit 3: again 3 of the 4 sequences of final two digits are possible, so 3*2=6 solutions
first digit 4: 2 of the 4 sequences of final two digits are possible, so 2*2=4 solutions<br>
The total number of 5-digit numbers using these digits that are divisible by 4 is 6+2+6+4 = 18<br>
The probability that a 5-digit number formed using these digits is divisible by 4 is 18/96 = 3/16.