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Difference between revisions of "2020 AIME I Problems/Problem 7"

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== Problem ==
 
== Problem ==
  
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== Solution ==
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== Solution 1 ==
 
We will be selecting girls, but <i>not</i> selecting boys. We claim that the amount of girls selected and the amount of guys not selected adds to <math>12</math>. This is easy to see: if <math>k</math> women were chosen, then <math>k + (11 - k + 1) = 12</math>. Therefore, we simply take <math>\binom{23}{12} \implies \boxed{081}</math>. ~awang11's sol
 
We will be selecting girls, but <i>not</i> selecting boys. We claim that the amount of girls selected and the amount of guys not selected adds to <math>12</math>. This is easy to see: if <math>k</math> women were chosen, then <math>k + (11 - k + 1) = 12</math>. Therefore, we simply take <math>\binom{23}{12} \implies \boxed{081}</math>. ~awang11's sol
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== Solution 2 (Bash) ==
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We casework on the amount of men on the committee.
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If there are no men in the committee, there are <math>\dbinom{12}{1}</math> ways to pick the women on the committee, for a total of <math>\dbinom{11}{0} \cdot \dbinom{12}{1}</math>. Notice that <math>\dbinom{11}{0}</math> is equal to <math>\dbinom{11}{11}</math>, 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 <math>\dbinom{12}{12}</math>. Therefore, these cases can be combined to <cmath>\dbinom{11}{0} \cdot (\dbinom{12}{1} + \dbinom{12}{12})</cmath> Since <math>\dbinom{12}{12} = \dbinom{12}{0}</math>, and <math>\dbinom{12}{0} + \dbinom{12}{1} = \dbinom{13}{1}</math>, we can further simplify this to <cmath>\dbinom{11}{0} \cdot \dbinom{13}{1}</cmath>
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All other cases proceed similarly. For example, the case with one men or ten men is equal to <math>\dbinom{11}{1} \cdot \dbinom{13}{2}</math>. Now, if we factor out a <math>13</math>, then all cases except the first two have a factor of <math>121</math>, so we can factor this out too to make our computation slightly easier. The first two cases (with <math>13</math> factored out) give <math>1+66=67</math>, and the rest gives <math>121(10+75+270+504) = 103,939</math>. Adding the <math>67</math> gives <math>104,006</math>. Now, we can test for prime factors. We know there is a factor of <math>2</math>, and the rest is <math>52,003</math>. We can also factor out a <math>7</math>, for <math>7,429</math>, and the rest is <math>17 \cdot 19 \cdot 23</math>. Adding up all the prime factors gives <math>2+7+13+17+19+23 = \boxed{081}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:17, 12 March 2020

Note: Please do not post problems here until after the AIME.

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 $12$. This is easy to see: if $k$ women were chosen, then $k + (11 - k + 1) = 12$. Therefore, we simply take $\binom{23}{12} \implies \boxed{081}$. ~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 $\dbinom{12}{1}$ ways to pick the women on the committee, for a total of $\dbinom{11}{0} \cdot \dbinom{12}{1}$. Notice that $\dbinom{11}{0}$ is equal to $\dbinom{11}{11}$, 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 $\dbinom{12}{12}$. Therefore, these cases can be combined to \[\dbinom{11}{0} \cdot (\dbinom{12}{1} + \dbinom{12}{12})\] Since $\dbinom{12}{12} = \dbinom{12}{0}$, and $\dbinom{12}{0} + \dbinom{12}{1} = \dbinom{13}{1}$, we can further simplify this to \[\dbinom{11}{0} \cdot \dbinom{13}{1}\]

All other cases proceed similarly. For example, the case with one men or ten men is equal to $\dbinom{11}{1} \cdot \dbinom{13}{2}$. Now, if we factor out a $13$, then all cases except the first two have a factor of $121$, so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$, and the rest gives $121(10+75+270+504) = 103,939$. Adding the $67$ gives $104,006$. Now, we can test for prime factors. We know there is a factor of $2$, and the rest is $52,003$. We can also factor out a $7$, for $7,429$, and the rest is $17 \cdot 19 \cdot 23$. Adding up all the prime factors gives $2+7+13+17+19+23 = \boxed{081}$.

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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