Question 397256
case 1: when 1 at first place 
           no. of ways to put even digit at third position = 5
           no. of ways to put any remaining digit at second position = 8 
           (except 1 and digit at  third position)
             
          total no. of possible numbers = 5*8 = 40



case 2: when 3 at first place, similar to previous condition

total no. of possible numbers = 5*8 = 40



case 3: when 2 at first place, 
         no. of ways to put even digit at third position = 4  (except 2)
           no. of ways to put any remaining digit at second position = 8 
           (except 1 and digit at  third position)      

   total no. of possible numbers = 4*8 = 32



total  = 40+40+32 = 112


try to understand the concept .........

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Mistake in your solution...
in case 1 (when 0 at last place)
first position can be filled by 3 ways (1,2 or 3).