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

Revision as of 19:19, 16 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 ${4 \choose 4}=1$ 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

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

@@ Line 16: / Line 16: @@
 ==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>{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
-<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}{4 \choose 4}=\frac{16!}{(4!)^5}</cmath>
 ways.

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

Page

Toolbox

Search