2020 AIME I Problems/Problem 7
Contents
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 1
Let be the number of women selected. Then, the number of men not selected is . Note that the sum of the number of women selected and the number of men not selected is constant at . Each combination of women selected and men not selected corresponds to a committee selection. Since choosing 12 individuals from the total of 23 would give women and men, the number of committee selections is . The answer is . ~awang11's sol
Solution 2 (Bash)
We casework on the amount of men on the committee.
If there are no men in the committee, there are ways to pick the women on the committee, for a total of . Notice that is equal to , so the case where no men are picked can be grouped with the case where all men are picked. When all men are picked, all women must also be picked, for a total of . Therefore, these cases can be combined to Since , and , we can further simplify this to
All other cases proceed similarly. For example, the case with one men or ten men is equal to . Now, if we factor out a , then all cases except the first two have a factor of , so we can factor this out too to make our computation slightly easier. The first two cases (with factored out) give , and the rest gives . Adding the gives . Now, we can test for prime factors. We know there is a factor of , and the rest is . We can also factor out a , for , and the rest is . Adding up all the prime factors gives .
Video Solution:
https://youtu.be/MVxsY8DwHVk ~ avn
Solution 3 (Vandermonde's identity)
Applying [1] by setting , , and , we obtain . ~Lcz
Short Proof
Consider the following setup: The dots to the left represent the men, and the dots to the right represent the women. Now, suppose we put a mark on people (the ). Those to the left of the dashed line get to be "in" on the committee if they have a mark. Those on the right side of the dashed line are already on the committee, but if they're marked they get forcibly evicted from it. If there were people marked on the left, there ends up being people not marked on the right. Circles represent those in the committee.
We have our bijection, so the number of ways will be .
~programjames1
Solution 4
Notice that the committee can consist of boys and girls. Summing over all possible gives Using the identity , and Pascal's Identity , we get Using the identity , this simplifies to so the desired answer is ~ktong
Solution 5 (Official MAA)
Select any club members. That group will have men and women, so the number of women in the club not selected in that group is . Thus, if the committee includes the men who were selected and the women who were not selected, the committee would have the correct number of men and women. Conversely, for every committee that could be formed with men and women, the men on this committee together with the women not on the committee comprise a subset of club members. ThusThe requested sum is
Video Solution by OmegaLearn
https://youtu.be/pGkLAX381_s?t=684
~ pi_is_3.14
Video Solution
(Solves using both methods - Casework and Vandermonde's Identity)
Video Solution
https://www.youtube.com/watch?v=fxlQMiElGFk&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=6 ~ MathEx
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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