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

(Solution)
(Video Solution 4 by sevenblade)
(17 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 +
{{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==
  
 
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 12: 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. 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. Therefore, there are
 
<cmath>\frac{16!}{(4!)^5}12^4</cmath>
 
<cmath>\frac{16!}{(4!)^5}12^4</cmath>
 
total possibilities.
 
total possibilities.
Line 24: Line 27:
  
 
~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino]
 
~[https://artofproblemsolving.com/community/user/1201585 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 <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
 +
 +
==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==
 +
{{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}}

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.

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

Fast Solution

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

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 $1$ 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. 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

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

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png