2024 AMC 12B Problems/Problem 16
- The following problem is from both the 2024 AMC 10B #22 and 2024 AMC 12B #16, so both problems redirect to this page.
Contents
- 1 Problem 16
- 2 Solution 1
- 3 Solution 2 (Multinomial Coefficients)
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution 1 by Pi Academy (In Less Than 2 Mins ⚡🚀)
- 7 Video Solution 2 by Innovative Minds
- 8 Video Solution 3 by SpreadTheMathLove
- 9 Video Solution 4 by sevenblade(standard approach!)
- 10 Fast Solution
- 11 See also
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 1
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
![\[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}=\frac{16!}{(4!)^5}\]](http://latex.artofproblemsolving.com/a/c/4/ac49c4c3de7fb6ce97b57384eb62bff557e28f7f.png)
ways.
In each committee, there are 
ways to choose the chairperson and secretary, so 
ways for all committees. 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 , 
.
Solution 2 (Multinomial Coefficients)
There are 
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 
ways to choose chairperson and secretary. Hence a total of 
~mathboy282
Solution 3
There will be ways to pick the chairperson of the first committee, then 
to pick the secretary, and lastly 
ways to pick the other two members of the first committee. Similarly, we can complete the rest of the terms as follows:
![\[\frac{(16)(15){14 \choose 2}(12)(11){10 \choose 2}(8)(7){6\choose 2}(4)(3){2\choose 2}}{4!}\]](http://latex.artofproblemsolving.com/4/6/6/4666f7d3e4dc368b6e5da13d14ff90e431b2d2a9.png)
We notice the numerator has at most 
, and the denominator has just . Thus, the value of in question is 
.
~lisztepos
Solution 4
We can convert the question into arranging people in a line of length .
First, arrange all people in a line. There are possible ways to do this. Then, divide the line into groups of , which represent the committees. Since the committees are indistinguishable, we divide by the number of ways to permute the groups, .
Within each group of , 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 
for each group. Thus, we divide by 
to account for all committees.
The total number of ways to assign people into committees with roles is given by:
![\[\frac{16!}{4! \cdot 2^4}.\]](http://latex.artofproblemsolving.com/f/7/0/f70ddacc0ba9e6d0fc4a4f03f06452bbd5f762d1.png)
(This is equivalent to Solution 1's 
.)
To find , we use Legendre's Formula to count the total power of in :
![\[\lfloor \frac{16}{3} \rfloor + \lfloor \frac{16}{9} \rfloor = 5 + 1 = 6.\]](http://latex.artofproblemsolving.com/9/4/c/94c73353fb3632012af18a8ef342c4a3fff1c374.png)
Each contributes one factor of , and since we divide by a total of 
times, we subtract from :
![\[6 - 1 = 5.\]](http://latex.artofproblemsolving.com/d/5/7/d575ba51817e657d9723242188d2a2e7f75c4b80.png)
Thus, the value of is 
.
~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
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
