Problem:
A club consisting of men and women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as member or as many as members. Let be the number of such committees that can be formed. Find the sum of the prime numbers that divide .
Solution:
Suppose we select any group of club members. Let this group contain men and women. Then the number of women in the club who are not in this group is .
Now, consider forming a committee consisting of the men from the selected group and the women not in the group. This gives a committee with exactly men and women.
Conversely, for any committee formed with men and women, we can pair the men on the committee with the women who are not on the committee. Together, these people form a group of club members.
Thus, there is a one-to-one correspondence between groups of club members and valid committees of men and women.
So,
The requested sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions