2021 AMC 10B Problems/Problem 22: Difference between revisions
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<math>\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542</math> | <math>\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542</math> | ||
==Solution== | ==Solution== | ||
Let our denominator be <math>(5!)^3</math>, so we consider all possible distributions. | |||
We use PIE (Principle of Inclusion and Exclusion) to count the successful ones. | |||
When we have at <math>1</math> box with all <math>3</math> balls the same color in that box, there are <math>_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3</math> ways for the distributions to occur (<math>_{5} C _{1}</math> for selecting one of the five boxes for a uniform color, <math>_{5} P _{1}</math> for choosing the color for that box, <math>4!</math> for each of the three people to place their remaining items). | |||
However, we overcounted those distributions where two boxes had uniform color, and there are <math>_{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3</math> ways for the distributions to occur (<math>_{5} C _{2}</math> for selecting two of the five boxes for a uniform color, <math>_{5} P _{2}</math> for choosing the color for those boxes, <math>3!</math> for each of the three people to place their remaining items). | |||
Again, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth. | |||
Our success by PIE is <cmath>_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3 - _{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3 + _{5} C _{3} \cdot _{5} P _{3} \cdot (2!)^3 - _{5} C _{4} \cdot _{5} P _{4} \cdot (1!)^3 + _{5} C _{5} \cdot _{5} P _{5} \cdot (0!)^3 = 120 \cdot 2556.</cmath> | |||
<cmath>\frac{120 \cdot 2556}{120^3}=\frac{71}{400},</cmath> yielding an answer of <math>471</math>. | |||
{{AMC10 box|year=2021|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2021|ab=B|num-b=21|num-a=23}} | ||
Revision as of 00:09, 12 February 2021
Problem
Ang, Ben, and Jasmin each have 
blocks, colored red, blue, yellow, white, and green; and there are empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives 
blocks all of the same color is 
, where 
and 
are relatively prime positive integers. What is 
Solution
Let our denominator be 
, so we consider all possible distributions.
We use PIE (Principle of Inclusion and Exclusion) to count the successful ones.
When we have at 
box with all balls the same color in that box, there are 
ways for the distributions to occur (
for selecting one of the five boxes for a uniform color, 
for choosing the color for that box, 
for each of the three people to place their remaining items).
However, we overcounted those distributions where two boxes had uniform color, and there are 
ways for the distributions to occur (
for selecting two of the five boxes for a uniform color, 
for choosing the color for those boxes, 
for each of the three people to place their remaining items).
Again, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth.
Our success by PIE is ![\[_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3 - _{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3 + _{5} C _{3} \cdot _{5} P _{3} \cdot (2!)^3 - _{5} C _{4} \cdot _{5} P _{4} \cdot (1!)^3 + _{5} C _{5} \cdot _{5} P _{5} \cdot (0!)^3 = 120 \cdot 2556.\]](http://latex.artofproblemsolving.com/a/5/9/a59888bd1785744fa33e206eac015cdc07be6a66.png)
![\[\frac{120 \cdot 2556}{120^3}=\frac{71}{400},\]](http://latex.artofproblemsolving.com/e/8/7/e87883620d06649fc386896ce567597f3e0fcbbd.png)
yielding an answer of 
.
| 2021 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
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