1) select the third digit
2) arrange the 3 digits
3) remove invalid cases
4) remove repeated cases
1) no of ways of selecting the third digit (any one from 0 to 9)= 10C1 = 10
2) no. of ways of arrangement of 3 digits = 3!
So total arrangements = 10*3! = 60
3) invalid cases are the one where 0 is the first digit. There are only 2 such numbers ( 026 and 062 )
invalid cases = 2
4) cases are repeated when the third digit is either 2 or 6
For each of the digits, the repeated cases are 3!/2! = 3
Hence total repeat cases for the 2 digits = 2 * 3 = 6
Hence required number = 60-2-6 = 52.
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