Difference between revisions of "2021 Fall AMC 12A Problems/Problem 18"
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==Solution 1 (Multinomial Coefficients)== | ==Solution 1 (Multinomial Coefficients)== | ||
− | For simplicity purposes, we assume that the balls | + | For simplicity purposes, we assume that the balls and the bins are both distinguishable. |
− | + | Recall that there are <math>5^{20}</math> ways to distribute <math>20</math> balls into <math>5</math> bins. We have | |
− | <cmath>p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{ | + | <cmath>p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.</cmath> Therefore, the answer is <cmath>\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.</cmath> |
+ | ~MRENTHUSIASM ~Jesshuang | ||
− | + | ==Solution 2 == | |
+ | For simplicity purposes, we assume that the balls and the bins are both distinguishable. | ||
− | + | Let <math>q=\frac{x}{a},</math> where <math>a</math> is the total number of combinations and <math>x</math> is the number of cases where every bin ends up with <math>4</math> balls. | |
− | + | We can take <math>1</math> ball from one bin and place it in another bin so that some bin ends up with <math>3</math> balls, another with <math>5</math> balls, and the other three with <math>4</math> balls each. Note that one configuration of <math>4{-}4{-}4{-}4{-}4</math> corresponds to <math>5\cdot4\cdot4=80</math> configurations of <math>3{-}5{-}4{-}4{-}4.</math> On the other hand, one configuration of <math>3{-}5{-}4{-}4{-}4</math> corresponds to <math>5</math> configurations of <math>4{-}4{-}4{-}4{-}4.</math> | |
− | + | Therefore, we have <cmath>p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},</cmath> from which <math>\frac{p}{q} = \boxed{\textbf{(E)}\ 16}.</math> | |
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~Hoju | ~Hoju | ||
==Solution 3 (Binomial Coefficients) == | ==Solution 3 (Binomial Coefficients) == | ||
− | Since both of the | + | Since both of the cases will have <math>3</math> bins with <math>4</math> balls in them, we can leave those out. There are <math>2 \cdot \binom {5}{2} = 20</math> ways to choose where to place the <math>3</math> and the <math>5</math>. After that, there are <math>\binom {8}{3} = 56</math> ways to put the <math>3</math> and <math>5</math> balls being put into the bins. For the <math>4,4,4,4,4</math> case, after we canceled the <math>4,4,4</math> out, we have <math>\binom {8}{4} = 70</math> ways to put the <math>4</math> balls inside the bins. Therefore, we have <math>\frac {56\cdot 20}{70}</math> which is equal to <math>8 \cdot 2 = \boxed{\textbf{(E)}\ 16}</math>. |
~Arcticturn | ~Arcticturn | ||
− | ==Solution 4 (Set Theory) == | + | ==Solution 4 (Set Theory / Graph Theory) == |
− | Construct the set <math>A</math> consisting of all possible <math>3-5-4-4-4</math> bin configurations, and construct set <math>B</math> consisting of all possible <math>4-4-4-4-4</math> configurations. If we let <math>N</math> be the total number of configurations possible, it's clear we want to solve for <math>\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}</math>. | + | Construct the set <math>A</math> consisting of all possible <math>3{-}5{-}4{-}4{-}4</math> bin configurations, and construct set <math>B</math> consisting of all possible <math>4{-}4{-}4{-}4{-}4</math> configurations. If we let <math>N</math> be the total number of configurations possible, it's clear we want to solve for <math>\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}</math>. |
− | Consider drawing an edge between an element in <math>A</math> and an element in <math>B</math> if it is possible to reach one configuration from the other by moving a single ball ( | + | Consider drawing an edge between an element in <math>A</math> and an element in <math>B</math> 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 <math>A</math>, we may choose one of the <math>5</math> balls in the <math>5</math>-bin and move it to the <math>3</math>-bin to get a valid element in <math>B</math>. This implies the number of edges is <math>5|A|</math>. | |
− | On the other hand for any element in <math>B</math>, we may choose one of the <math>20</math> balls and move it to one of the other <math>4</math> bins to get a valid element in <math>A</math>. This implies the number of edges is <math>80|B|</math>. | + | On the other hand, for any element in <math>B</math>, we may choose one of the <math>20</math> balls and move it to one of the other <math>4</math>-bins to get a valid element in <math>A</math>. This implies the number of edges is <math>80|B|</math>. |
− | + | We equate the expressions to get <math>5|A| = 80|B|</math>, from which <math>\frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}</math>. | |
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/mIJ8VMuuVvA?t=220 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Video Solution by Mathematical Dexterity== | ==Video Solution by Mathematical Dexterity== | ||
Line 49: | Line 51: | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://YouTube.com/watch?v=bvd2VjMxiZ4 | https://YouTube.com/watch?v=bvd2VjMxiZ4 | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/TOSHQPb7vaM | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Latest revision as of 12:12, 10 November 2023
- 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. We have
Therefore, the answer is
~MRENTHUSIASM ~Jesshuang
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.