A dice is thrown six times, it being known that each time a different digit is shown. The probability that a sum of 12 will be obtained in the first three throws is

Since it is known that each time a different digit appears, total number of cases is 6!The sum in the first three throws is 12 if we get (1, 5, 6), (2,4,6) or (3 , 4, 5) in any order.So, number of cases for first three throws = 3 x 3!The remaining three throws have 3! ways. ∴ Number of favourable cases = 3 x 3! x 3!∴ Required probability=3×3!×3!6!=320