2020 AIME I Problems/Problem 7
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[hide]Problem
Solution 1
We will be selecting girls, but not selecting boys. We claim that the amount of girls selected and the amount of guys not selected adds to 
. This is easy to see: if 
women were chosen, then 
. Therefore, we simply take 
. ~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 females must also be picked, for a total of 
. Therefore, these cases can be combined to ![\[\dbinom{11}{0} \cdot (\dbinom{12}{1} + \dbinom{12}{12})\]](http://latex.artofproblemsolving.com/4/2/0/420e6b0a19f66ceffe7255a2db15f6bd7d01a564.png)
Since 
, and 
, we can further simplify this to ![\[\dbinom{11}{0} \cdot \dbinom{13}{1}\]](http://latex.artofproblemsolving.com/1/2/a/12a522ac0cf7513e837f3e2459d5d3aac4e161b9.png)
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 
.
See Also
| 2020 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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