2021 Fall AMC 12A Problems/Problem 18
- The following problem is from both the 2021 Fall AMC 10A #21 and 2021 Fall AMC 12A #18, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Multinomial Coefficients)
- 3 Solution 2
- 4 Solution 3 (Binomial Coefficients)
- 5 Solution 4 (Set Theory / Graph Theory)
- 6 Video Solution by OmegaLearn
- 7 Video Solution by Mathematical Dexterity
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution by TheBeautyofMath
- 10 See Also
Problem
Each of the balls is tossed independently and at random into one of the bins. Let be the probability that some bin ends up with balls, another with balls, and the other three with balls each. Let be the probability that every bin ends up with balls. What is ?
Solution 1 (Multinomial Coefficients)
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Recall that there are ways to distribute balls into bins. For we choose one of the bins to have balls and another one of the bins to have balls. We get Therefore, the answer is ~MRENTHUSIASM ~Jesshuang ~mathboy282
Solution 2
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Let where is the total number of combinations and is the number of cases where every bin ends up with balls.
We can take ball from one bin and place it in another bin so that some bin ends up with balls, another with balls, and the other three with balls each. Note that one configuration of corresponds to configurations of On the other hand, one configuration of corresponds to configurations of
Therefore, we have from which
~Hoju
Solution 3 (Binomial Coefficients)
Since both of the cases will have bins with balls in them, we can leave those out. There are ways to choose where to place the and the . After that, there are ways to put the and balls being put into the bins. For the case, after we canceled the out, we have ways to put the balls inside the bins. Therefore, we have which is equal to .
~Arcticturn
Solution 4 (Set Theory / Graph Theory)
Construct the set consisting of all possible bin configurations, and construct set consisting of all possible configurations. If we let be the total number of configurations possible, it's clear we want to solve for .
Consider drawing an edge between an element in and an element in if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.
For any element in , we may choose one of the balls in the -bin and move it to the -bin to get a valid element in . This implies the number of edges is .
On the other hand, for any element in , we may choose one of the balls and move it to one of the other -bins to get a valid element in . This implies the number of edges is .
We equate the expressions to get , from which .
Video Solution by OmegaLearn
https://youtu.be/mIJ8VMuuVvA?t=220
~ pi_is_3.14
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=Lu6eSvY6RHE
Video Solution by Punxsutawney Phil
https://YouTube.com/watch?v=bvd2VjMxiZ4
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.