1. Given a 6x5 array of unique elements, how many ways can three elements be selected so that no two are in the same row or column? 2. Fundamental Counting Principle? 3. This question was a sample question in the Actuary P1 exam prep I am reviewing. I thought it would be easy enough to use Counting Principle to determine the solution. So, the set of your first choice has 30 elements. Since this eliminates exactly one row and column, the set of your second choice only contains the remaining 4x5 array with 20 elements. And so on to get 12 elements for the third choice. Using FCP to determine the total number of combinations SHOULD therefore yield a solution of 7200. However, the provided solution for this question from the exam prep is different. What have I overlooked?