Let S be the set of permutations of (1,2,3,4,5,6) whose first term is not 1. If we choose a permutation at random from S, what is the probability that the third term is equal to 3, and the fourth term is equal to 4?
Possible sets =
1 can appear in any of the 5 other positions and the other elements can be arranged in 5! = 120 ways
So 5 *120 = 600 sets
In the sets of interest :
1 can appear in positions 2, 5 or 6 = 3 ways
And 2, 5 and 6 can be arranged in 3! = 6 ways in the remaining positions
So.... 3 * 6 = 18 sets
Probability = 18/600 = 3/100