There is a total of 8 students, half of them men and half of them women, and we are asked if we choose 4 of them, what is the probability that 2 of the 4 are men and two out of the four are women

Given the complexity of the problem, it seems best to view this as a counting problem and as order doesn't matter (i.e. selecting male #1 and male #2 is the same as selecting male number two and male number one), we can use combinations.

Working through the problem logically, we can break it up into two parts; first considering choosing the men, then consider choosing the women.

Selecting two out of the four men can be written as 4C2 while selecting two women out of the four in the group is similarly written as 4C2. By multiplying the resulting numbers, we get the total number of selections that could possibly occur. Then by dividing by the total number of possible choices, 8C4, we can find the probability of such a selection occurring.

(4C2)(4C2)/(8C4) = [4!/(2!)(2!) x 4!/(2!)(2!)] / [8!/(4!)(4!)] = (6)(6)/(70) = 36/70 = 18/35 ---->Answer Choice A