The wording of the question could have been better. Regardless, I am pretty sure the intent was to compute the possible permutations, which was my answer. Otherwise, the number of participants in the two categories would not have been presented.
Quote:
Originally Posted by b0bd0herty
Actually, there are only 3 awards so THOSE 3 awards can only be present in 6 possible arrangements. That makes 6 and 6.
|