Four die are thrown simultaneously. The probability that 4 and 3 appear on two of the die given that 5 and 6 have appeared on other two die is

Given that 5 and 6 have appeared on two of the dice, the sample space reduces to 64 - 2 x 54 + 44 (inclusion-exclusion principle). Also, the number of favorable cases are 4! = 24. So, the required probability is 24302=12151