Question 1097086
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0)  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.



    In order for the number is divisible by 4, the last two digits must be 

        04, 12, 24, 32, 20 and/or 40.


1)  If the last two digits (in the 4-th and 5-th positions, counting from the left)
    
    are 04, then in the 123-positions we have all permutations of the three digits 

    1, 2 and 3, which gives us 3! = 6 numbers.



2)  If the last two digits (in the 4-th and 5-th positions, counting from the left)
    
    are 12, then in the 123-positions we have permutations of the three digits 

    0, 3 and 4, with prohibited 0 in the first position, which gives us 2*2*1 = 4  numbers.



3)  If the last two digits (in the 4-th and 5-th positions, counting from the left)
    
    are 24, then in the 123-positions we have permutations of the three digits 

    0, 1 and 3, with prohibited 0 in the first position, which gives us again 2*2*1 = 4  numbers.



4)  If the last two digits (in the 4-th and 5-th positions, counting from the left)
    
    are 32, then in the 123-positions we have permutations of the three digits 

    0, 1 and 4, with prohibited 0 in the first position, which gives us again 2*2*1 = 4  numbers.



5)  If the last two digits (in the 4-th and 5-th positions, counting from the left)
    
    are 20, then in the 123-positions we have all permutations of the three digits 

    1, 3 and 4, which gives us 3! = 6 numbers.



6)  If the last two digits (in the 4-th and 5-th positions, counting from the left)
    
    are 40, then in the 123-positions we have all permutations of the three digits 

    1, 2 and 3, which gives us 3! = 6 numbers.



7)  Thus we have 96 allowed 5-digit numbers in all;  of them  exactly 6 + 4 + 4 + 4 + 6 + 6 = 30 are divisible by 4.


    Therefore, the probability under the question is  {{{30/96}}} = {{{5/16}}}.
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