Difference between revisions of "2024 AMC 12A Problems/Problem 16"
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There are a total of 3!(6c3)(3c2)(1c1) ways in doing so | There are a total of 3!(6c3)(3c2)(1c1) ways in doing so | ||
This the answer 3!(6!/1!2!3!)/(12!/4!4!4!)=4/385 | This the answer 3!(6!/1!2!3!)/(12!/4!4!4!)=4/385 | ||
| − | So the answer is <math>\boxed{\textbf{(C) }389}</math> | + | So the answer is <math>\boxed{\textbf{(C) }389}</math> |
| + | ~lptoggled | ||
| + | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=15|num-a=17}} | {{AMC12 box|year=2024|ab=A|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:46, 8 November 2024
Problem
A set of 
tokens ---- 
red, 
white, 
blue, and 
black ---- is to be distributed at random to game players, 
tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as 
, where 
and 
are relatively prime positive integers. What is 
?
Solution 1 (12fact bash)
We have 
total possible arrangements of distinct tokens. If we imagine the first tokens of our arrangement go to the first player, the next go to the second, and the final go to the third, then we can view this problem as counting the number of valid arrangements.
Firstly, the tokens are not all distinct, so we multiply by 
, 
, 
, and 
to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.
Letting 
denote red, 
denote white, 
denote blue, and 
denote black, then our arrangement must be something like 
. The three players are arbitrary, so we multiply by ; then, the player who gets the reds has 
possible arrangements, the player who gets the whites has 
possibilities, and the player who gets the blacks has 
possibilities. Our total on top is thus 
, and the denominator is . Firstly, we have the in the numerator cancel out part of the denominator; we thus have the following:
![\[\dfrac{3\cdot2\cdot2\cdot3\cdot2\cdot4\cdot6\cdot4}{12\cdot11\cdot10\cdot9\cdot8\cdot7}=\dfrac{2^83^3}{2^63^35\cdot7\cdot11}=\dfrac4{385}.\]](http://latex.artofproblemsolving.com/4/6/2/4626567e9829f4d9c20a59621b51fa34eec8c553.png)
Our answer is 
~Technodoggo
Solution 2
Assume all of them are distinct even when some have the same color,
Total possibility
Next, assume that all the token are already in 3 different groups (3! Ways to do so)
We then split the 6 black token into these 3 different groups
There are a total of 3!(6c3)(3c2)(1c1) ways in doing so
This the answer 3!(6!/1!2!3!)/(12!/4!4!4!)=4/385
So the answer is 
~lptoggled
See also
| 2024 AMC 12A (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. 
