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Difference between revisions of "2024 AMC 12B Problems/Problem 16"

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{{duplicate|[[2024 AMC 10B Problems/Problem 22|2024 AMC 10B #22]] and [[2024 AMC 12B Problems/Problem 16|2024 AMC 12B #16]]}}
 
==Problem 16==
 
==Problem 16==
  
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[[2024 AMC 12B Problems/Problem 16|Solution]]
 
[[2024 AMC 12B Problems/Problem 16|Solution]]
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==Solution==
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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
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<cmath>\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}=\frac{16!}{(4!)^5}</cmath>
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ways.
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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
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<cmath>\frac{16!}{(4!)^5}12^4</cmath>
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total possibilities.
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Since <math>16!</math> contains <math>6</math> factors of <math>3</math>, <math>(4!)^5</math> contains <math>5</math>, and <math>12^4</math> contains <math>4</math>, <math>r=6-5+4=\boxed{\textbf{(A) }5}</math>.
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~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino]
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==Video Solution by Innovative Minds==
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https://youtu.be/HMPHdBiaYQc
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==See also==
 
==See also==
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{{AMC10 box|year=2024|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2024|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2024|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:29, 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.

Problem 16

A group of $16$ people will be partitioned into $4$ indistinguishable $4$-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^{r}M$, where $r$ and $M$ are positive integers and $M$ is not divisible by $3$. What is $r$?

$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

Solution

Solution

There are ${16 \choose 4}$ ways to choose the first committee, ${12 \choose 4}$ ways to choose the second, ${8 \choose 4}$ for the third, and $1$ for the fourth. Since the committees are indistinguishable, we need to divide the product by $4!$. Thus the $16$ people can be grouped in \[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}=\frac{16!}{(4!)^5}\] ways.

In each committee, there are $4 \cdot 3=12$ ways to choose the chairperson and secretary, so $12^4$ ways for all $4$ committees. Note that we do not divide by $4!$ 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 \[\frac{16!}{(4!)^5}12^4\] total possibilities.

Since $16!$ contains $6$ factors of $3$, $(4!)^5$ contains $5$, and $12^4$ contains $4$, $r=6-5+4=\boxed{\textbf{(A) }5}$.

~kafuu_chino

Video Solution by Innovative Minds

https://youtu.be/HMPHdBiaYQc

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

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