2023 AIME I Problem 1

The $14$ positions split into $7$ diametrically opposite pairs, and only the set of positions the men occupy matters.

Count sets where no pair has two men: choose $5$ of the $7$ pairs to hold a man, then a position in each; compare with $\binom{14}{5}.$

The $14$ positions split into $7$ diametrically opposite pairs. Only the set of positions occupied by the men matters, and all $\binom{14}{5} = 2002$ five-element sets are equally likely. Every man stands opposite a woman exactly when no pair contains two men, so choose which $5$ of the $7$ pairs contain a man ($\binom{7}{5} = 21$ ways) and which position of each chosen pair the man occupies ($2^5 = 32$ ways), for $21 \cdot 32 = 672$ favorable sets.

The probability is $\frac{672}{2002} = \frac{48}{143},$ so $m + n = 48 + 143 = 191.$