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. Note that we do not divide by <math>4!</math> here since the choosing of the two positions for each committee are independent of each other (not sure how to explain this). Therefore, there are | 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. Note that we do not divide by <math>4!</math> here since the choosing of the two positions for each committee are independent of each other (not sure how to explain this). 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 01:21, 14 November 2024
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 ?
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. Note that we do not divide by here since the choosing of the two positions for each committee are independent of each other (not sure how to explain this). 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 , 
.
See also
| 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. 
