Question 341008
Digits are    0,1,3,5,6,8

 we have to fill four places  ABCD, D is unit digit, B is tenth digit and so on..

    number must be greater than 6000 so, 

no. of ways to fill position of A  is  2     (6 or 8) 

case 1: when 6 is at A (first digit)
             
           no. of ways to fill the position D =  3 (1,3 or 5)
           no. of ways to fill the position B = 4  (only 4 digit remaining)
             no. of ways to fill the position c= 2   (only 3 digit )
          
        total no. of ways to form odd integer starting from 6 = 3*4*2= 24

case 2: when 8 is at A (first digit)
                     
           similarly, no. of ways to fill the position D = 3
                        no. of ways to fill the position B = 4 
                         no. of ways to fill the position C = 3

       total no. of ways to form odd integer starting from 8 = 3*4*2 =24


    total no. of four digit odd integers greater than 6000 = 24+24 = 48