Let S be the set of permutations of (1,2,3,4,5,6) whose first term is a prime.
If we choose a permutation at random from S, what is the probability that the third term is equal to 2?
+23186 The first term can be either 2, 3, or 5.
If the first term is 2, there are 5! = 5 x 4 x 3 x 2 x 1 = 120 ways to permutate the other terms.
Similarly, there are 120 ways if the first term is a 3 and there are 120 ways if the first term is a 5.
Therefore there is a total of 360 possible permutations for the denominator of the probability.
If the third term is a 2 and the first term is a 3, there are 4 x 3 x 2 x 1 = 24 ways.
Similarly, if the third term is a 2 and the first term is a 5, there are 24 ways.
Adding, we get 48 ways for the numerator of the probability.
The probability will be 24 / 360.
