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
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 are indistinguishable and the bins are distinguishable.
Let be the number of ways to distribute balls into bins. We have Therefore, the answer is
Remark
By the stars and bars argument, we get ~MRENTHUSIASM
Solution 2 (Binomial Coefficients)
For simplicity purposes, the balls are indistinguishable and the bins are distinguishable.
Let be equal to where is the total number of combinations and is the number of cases where every bin ends up with balls.
Notice that 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. We have Therefore, we get from which
~Hoju
Solution 3 (Binomial Coefficients)
Since both of the boxes will have boxes 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 boxes. For the case, after we canceled the out, we have ways to put the balls inside the boxes. Therefore, we have which is equal to .
~Arcticturn
Solution 4 (Set 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.
From any element in , we may take one of the balls in the 5-bin and move it to the 3-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 .
Since they must be equal, then .
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=Lu6eSvY6RHE
Video Solution by Punxsutawney Phil
https://YouTube.com/watch?v=bvd2VjMxiZ4
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.