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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>?
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<math>
+
 
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\textbf{(A) }5 \qquad
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<math>\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad</math>
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\textbf{(B) }6 \qquad
 
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\textbf{(C) }7 \qquad
 
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\textbf{(D) }8 \qquad
 
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\textbf{(E) }9 \qquad</math>
 
  
 
[[2024 AMC 12B Problems/Problem 16|Solution]]
 
[[2024 AMC 12B Problems/Problem 16|Solution]]
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−
==Fast Solution==
 
−
https://www.youtube.com/watch?v=jPTL8hf0Ur0&t=1s
 
  
 
==Solution 1==
 
==Solution 1==
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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
+
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>{4 \choose 4}=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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<cmath>\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}=\frac{16!}{(4!)^5}</cmath>
 
ways.
 
ways.
  
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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:
 
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>
 
<cmath>\frac{(16)(15){14 \choose 2}(12)(11){10 \choose 2}(8)(7){6\choose 2}(4)(3){2\choose 2}}{4!}</cmath>
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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 \boxed{\textbf{(A)}\ 5}.
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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
 
~lisztepos
 +
 +
==Solution 4 == 
 +
We can convert the question into arranging people in a line of length <math>16</math>.
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 +
First, arrange all <math>16</math> people in a line. There are <math>16!</math> possible ways to do this. Then, divide the line into <math>4</math> groups of <math>4</math>, which represent the committees. Since the committees are indistinguishable, we divide by the number of ways to permute the <math>4</math> groups, <math>4!</math>. 
 +
 +
Within each group of <math>4</math>, the first two positions in the line are assigned as the chairperson and the secretary, while the remaining two are ordinary members. Since the roles of chairperson and secretary are distinct, but the order of the remaining two members does not matter, we divide by <math>2</math> for each group. Thus, we divide by <math>2^4</math> to account for all <math>4</math> committees. 
 +
 +
The total number of ways to assign people into committees with roles is given by:
 +
<cmath>
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\frac{16!}{4! \cdot 2^4}.
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</cmath> 
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(This is equivalent to Solution 1's <math>\frac{16!}{(4!)^5} \cdot 12^4</math>.)
 +
 +
To find <math>r</math>, we use Legendre's Formula to count the total power of <math>3</math> in <math>16!</math>: 
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<cmath>
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\lfloor \frac{16}{3} \rfloor + \lfloor \frac{16}{9} \rfloor = 5 + 1 = 6.
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</cmath> 
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Each <math>4!</math> contributes one factor of <math>3</math>, and since we divide by <math>4!</math> a total of <math>1</math> times, we subtract <math>1</math> from <math>6</math>: 
 +
<cmath>
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6 - 1 = 5.
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</cmath> 
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Thus, the value of <math>r</math> is <math>\boxed{5}</math>.
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~ryk
  
 
==Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)==
 
==Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)==
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https://www.youtube.com/watch?v=24EZaeAThuE
 
https://www.youtube.com/watch?v=24EZaeAThuE
  
 +
==Video Solution 4 by sevenblade(standard approach!)==
 +
https://www.youtube.com/watch?v=5BXclh_DLEg
 +
 +
==Fast Solution==
 +
https://www.youtube.com/watch?v=jPTL8hf0Ur0&t=1s
  
 
==See also==
 
==See also==

Latest revision as of 19:49, 17 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 1

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

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 $\binom{16}{4,4,4,4}$ 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 $4 \cdot 3$ ways to choose chairperson and secretary. Hence a total of $\lfloor{\frac{16}{3}\rfloor}+\lfloor{\frac{16}{9}\rfloor}+4-5=5.$

~mathboy282

Solution 3

There will be $16$ ways to pick the chairperson of the first committee, then $15$ to pick the secretary, and lastly ${14 \choose 2}$ ways to pick the other two members of the first committee. Similarly, we can complete the rest of the terms as follows: \[\frac{(16)(15){14 \choose 2}(12)(11){10 \choose 2}(8)(7){6\choose 2}(4)(3){2\choose 2}}{4!}\] We notice the numerator has at most $3^6$, and the denominator has just $3$. Thus, the value of $r$ in question is $\boxed{\textbf{(A)}\ 5}$.

~lisztepos

Solution 4

We can convert the question into arranging people in a line of length $16$.

First, arrange all $16$ people in a line. There are $16!$ possible ways to do this. Then, divide the line into $4$ groups of $4$, which represent the committees. Since the committees are indistinguishable, we divide by the number of ways to permute the $4$ groups, $4!$.

Within each group of $4$, the first two positions in the line are assigned as the chairperson and the secretary, while the remaining two are ordinary members. Since the roles of chairperson and secretary are distinct, but the order of the remaining two members does not matter, we divide by $2$ for each group. Thus, we divide by $2^4$ to account for all $4$ committees.

The total number of ways to assign people into committees with roles is given by: \[\frac{16!}{4! \cdot 2^4}.\] (This is equivalent to Solution 1's $\frac{16!}{(4!)^5} \cdot 12^4$.)

To find $r$, we use Legendre's Formula to count the total power of $3$ in $16!$: \[\lfloor \frac{16}{3} \rfloor + \lfloor \frac{16}{9} \rfloor = 5 + 1 = 6.\] Each $4!$ contributes one factor of $3$, and since we divide by $4!$ a total of $1$ times, we subtract $1$ from $6$: \[6 - 1 = 5.\] Thus, the value of $r$ is $\boxed{5}$. ~ryk

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

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

Fast Solution

https://www.youtube.com/watch?v=jPTL8hf0Ur0&t=1s

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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