Difference between revisions of "2021 Fall AMC 12A Problems/Problem 18"
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Since they must be equal, then <math>5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}</math>. | Since they must be equal, then <math>5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}</math>. | ||
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==Video Solution by Mathematical Dexterity== | ==Video Solution by Mathematical Dexterity== |
Revision as of 02:02, 26 November 2021
- 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.