Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 10A Problems/Problem 21"

m (→‎Solution 1 (Multinomial Numbers))
Line 21: Line 21:
  
 
~Arcticturn
 
~Arcticturn
 +
 +
==Solution 3 (30-second set-theoretic solution) ==
 +
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 (note this process is reversible). Let us consider the total number of edges drawn.
 +
 +
From any element in <math>A</math>, we may take one of the <math>5</math> balls in the 5-bin and move it to the 3-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>.
 +
 +
Since they must be equal, then <math>5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed {(E)16}</math>
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Revision as of 23:28, 22 November 2021

Problem

Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$

Solution 1 (Multinomial Coefficients)

For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable.

Let $d$ be the number of ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.\]

Remark

By the stars and bars argument, we get $d=\binom{20+5-1}{5-1}=\binom{24}{4}.$

~MRENTHUSIASM

Solution 2 (Simple)

Since both of the boxes will have $3$ boxes with $4$ balls in them, we can leave those out. There are $\binom {6}{3}$ = $20$ ways to choose where to place the $3$ and the $5$. After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the boxes. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4}$ = $70$ ways to put the $4$ balls inside the boxes. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2$ = $\boxed {(E)16}$

~Arcticturn

Solution 3 (30-second set-theoretic solution)

Construct the set $A$ consisting of all possible $3-5-4-4-4$ bin configurations, and construct set $B$ consisting of all possible $4-4-4-4-4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$.

Consider drawing an edge between an element in $A$ and an element in $B$ 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 $A$, we may take one of the $5$ balls in the 5-bin and move it to the 3-bin to get a valid element in $B$. This implies the number of edges is $5|A|$.

On the other hand for any element in $B$, we may choose one of the $20$ balls and move it to one of the other $4$ bins to get a valid element in $A$. This implies the number of edges is $80|B|$.

Since they must be equal, then $5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed {(E)16}$

Video Solution by Punxsutawney Phil

https://YouTube.com/watch?v=bvd2VjMxiZ4

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_21&oldid=165269"