Difference between revisions of "2024 AMC 12B Problems/Problem 16"
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| − | 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. | ||
Revision as of 12:22, 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
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
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
![\[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}=\frac{16!}{(4!)^5}\]](http://latex.artofproblemsolving.com/e/c/d/ecda11e2998d808c835dc388cbd7bb55be353a7a.png)
ways.
In each committee, there are 
ways to choose the chairperson and secretary, so 
ways for all committees. Therefore, there are
![\[\frac{16!}{(4!)^5}12^4\]](http://latex.artofproblemsolving.com/4/a/8/4a8497e088d9d2cfde86b5c38f6b05c3a097a053.png)
total possibilities.
Since 
contains 
factors of , 
contains 
, and contains , 
.
Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)
https://youtu.be/9ymwnHnTbDQ?feature=shared
~ Pi Academy
Video Solution by Innovative Minds
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
