Difference between revisions of "2024 AMC 12B Problems/Problem 16"
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A group of <math>16</math> people will be partitioned into <math>4</math> indistinguishable <math>4</math>-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as <math>3^{r}M</math>, where <math>r</math> and <math>M</math> are positive integers and <math>M</math> is not divisible by <math>3</math>. What is <math>r</math>? | A group of <math>16</math> people will be partitioned into <math>4</math> indistinguishable <math>4</math>-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as <math>3^{r}M</math>, where <math>r</math> and <math>M</math> are positive integers and <math>M</math> is not divisible by <math>3</math>. What is <math>r</math>? | ||
− | |||
<math> | <math> | ||
\textbf{(A) }5 \qquad | \textbf{(A) }5 \qquad | ||
Line 13: | Line 12: | ||
[[2024 AMC 12B Problems/Problem 16|Solution]] | [[2024 AMC 12B Problems/Problem 16|Solution]] | ||
− | ==Solution== | + | ==Fast Solution== |
+ | https://www.youtube.com/watch?v=jPTL8hf0Ur0&t=1s | ||
+ | |||
+ | ==Solution 1== | ||
There are <math>{16 \choose 4}</math> ways to choose the first committee, <math>{12 \choose 4}</math> ways to choose the second, <math>{8 \choose 4}</math> for the third, and <math>1</math> for the fourth. Since the committees are indistinguishable, we need to divide the product by <math>4!</math>. Thus the <math>16</math> people can be grouped in | There are <math>{16 \choose 4}</math> ways to choose the first committee, <math>{12 \choose 4}</math> ways to choose the second, <math>{8 \choose 4}</math> for the third, and <math>1</math> for the fourth. Since the committees are indistinguishable, we need to divide the product by <math>4!</math>. Thus the <math>16</math> people can be grouped in | ||
<cmath>\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}=\frac{16!}{(4!)^5}</cmath> | <cmath>\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}=\frac{16!}{(4!)^5}</cmath> | ||
ways. | ways. | ||
− | In each committee, there are <math>4 \cdot 3=12</math> ways to choose the chairperson and secretary, so <math>12^4</math> ways for all <math>4</math> committees | + | In each committee, there are <math>4 \cdot 3=12</math> ways to choose the chairperson and secretary, so <math>12^4</math> ways for all <math>4</math> committees. Therefore, there are |
<cmath>\frac{16!}{(4!)^5}12^4</cmath> | <cmath>\frac{16!}{(4!)^5}12^4</cmath> | ||
total possibilities. | total possibilities. | ||
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~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino] | ~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino] | ||
− | ==Video Solution by Innovative Minds== | + | ==Note== |
+ | |||
+ | This problem would be vague if not for answer choices. If this problem were given without answer choices, we would have another possible answer, 1 (which would arise if it is possible for the chairperson and secretary of the same committee to be the same person). We get this by replacing the 12^4 in the solution with 16^4. | ||
+ | |||
+ | ==Solution 2 (Multinomial Coefficients)== | ||
+ | There are <math>\binom{16}{4,4,4,4}</math> ways to choose the 4 committees. You have to divide by another 4! since the order of the committees does not matter. Furthermore, in each committee, you can have <math>4 \cdot 3</math> ways to choose chairperson and secretary. Hence a total of <math>\lfloor{\frac{16}{3}\rfloor}+\lfloor{\frac{16}{9}\rfloor}+4-5=5.</math> | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | There will be <math>16</math> ways to pick the chairperson of the first committee, then <math>15</math> to pick the secretary, and lastly <math>{14 \choose 2}</math> ways to pick the other two members of the first committee. Similarly, we can complete the rest of the terms as follows: | ||
+ | <cmath>\frac{(16)(15){14 \choose 2}(12)(11){10 \choose 2}(8)(7){6\choose 2}(4)(3){2\choose 2}}{4!}</cmath> | ||
+ | We notice the numerator has at most <math>3^6</math>, and the denominator has just <math>3</math>. Thus, the value of <math>r</math> in question is <math>\boxed{\textbf{(A)}\ 5}</math>. | ||
+ | |||
+ | ~lisztepos | ||
+ | |||
+ | ==Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)== | ||
+ | |||
+ | https://youtu.be/9ymwnHnTbDQ?feature=shared | ||
+ | |||
+ | ~ Pi Academy | ||
+ | |||
+ | ==Video Solution 2 by Innovative Minds== | ||
https://youtu.be/HMPHdBiaYQc | https://youtu.be/HMPHdBiaYQc | ||
+ | |||
+ | ==Video Solution 3 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=24EZaeAThuE | ||
+ | |||
+ | ==Video Solution 4 by sevenblade(standard approach!)== | ||
+ | https://www.youtube.com/watch?v=5BXclh_DLEg | ||
==See also== | ==See also== |
Revision as of 19:54, 14 November 2024
- The following problem is from both the 2024 AMC 10B #22 and 2024 AMC 12B #16, so both problems redirect to this page.
Contents
[hide]- 1 Problem 16
- 2 Fast Solution
- 3 Solution 1
- 4 Note
- 5 Solution 2 (Multinomial Coefficients)
- 6 Solution 3
- 7 Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)
- 8 Video Solution 2 by Innovative Minds
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution 4 by sevenblade(standard approach!)
- 11 See also
Problem 16
A group of people will be partitioned into indistinguishable -person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as , where and are positive integers and is not divisible by . What is ?
Fast Solution
https://www.youtube.com/watch?v=jPTL8hf0Ur0&t=1s
Solution 1
There are ways to choose the first committee, ways to choose the second, for the third, and for the fourth. Since the committees are indistinguishable, we need to divide the product by . Thus the people can be grouped in ways.
In each committee, there are ways to choose the chairperson and secretary, so ways for all committees. Therefore, there are total possibilities.
Since contains factors of , contains , and contains , .
Note
This problem would be vague if not for answer choices. If this problem were given without answer choices, we would have another possible answer, 1 (which would arise if it is possible for the chairperson and secretary of the same committee to be the same person). We get this by replacing the 12^4 in the solution with 16^4.
Solution 2 (Multinomial Coefficients)
There are ways to choose the 4 committees. You have to divide by another 4! since the order of the committees does not matter. Furthermore, in each committee, you can have ways to choose chairperson and secretary. Hence a total of
~mathboy282
Solution 3
There will be ways to pick the chairperson of the first committee, then to pick the secretary, and lastly ways to pick the other two members of the first committee. Similarly, we can complete the rest of the terms as follows: We notice the numerator has at most , and the denominator has just . Thus, the value of in question is .
~lisztepos
Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)
https://youtu.be/9ymwnHnTbDQ?feature=shared
~ Pi Academy
Video Solution 2 by Innovative Minds
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
Video Solution 4 by sevenblade(standard approach!)
https://www.youtube.com/watch?v=5BXclh_DLEg
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.