Problem:
Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is , where and are relatively prime positive integers. Find .
**Solution:**For simplicity, we consider two arrangements different even if they differ only by rotations or reflections. In this way, there are total arrangements without restrictions.
First, there are ways to choose the five diameters that will each contain a man and a woman.
Then, there are ways to assign five men to those chosen diameters (since there are 10 men total and we choose one for each diameter, reducing the number of available men by 2 each time because each man-woman pair occupies a full diameter).
Next, the remaining nine women can be arranged in the remaining seats in ways.
Putting everything together, the desired probability is:
\frac{\binom{7}{5} \cdot (10 \cdot 8 \cdot 6 \cdot 4 \cdot 2) \cdot 9!}
From which the answer is:
48 + 143 = \boxed
The problems on this page are the property of the MAA's American Mathematics Competitions