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Difference between revisions of "2017 AMC 8 Problems/Problem 20"

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==Problem 20==
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==Problem==
  
 
An integer between <math>1000</math> and <math>9999</math>, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?
 
An integer between <math>1000</math> and <math>9999</math>, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?
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==Solution==
 
==Solution==
  
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There are <math>5</math> options for the last digit, as the integer must be odd. The first digit now has <math>8</math> options left (it can't be <math>0</math> or the same as the last digit. The second digit also has <math>8</math> options left (it can't be the same as the first or last digit). Finally, the third digit has <math>7</math> options (it can't be the same as the three digits that are already chosen).  
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There are <math>5</math> options for the last digit as the integer must be odd. The first digit now has <math>8</math> options left (it can't be <math>0</math> or the same as the last digit). The second digit also has <math>8</math> options left (it can't be the same as the first or last digit). Finally, the third digit has <math>7</math> options (it can't be the same as the three digits that are already chosen).  
  
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Since there are <math>9000</math> total integers, out answer is <cmath>\frac{8 \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.</cmath>
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Since there are <math>9,000</math> total integers, our answer is <cmath>\frac{8 \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.</cmath>
  
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~nukelauncher
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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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https://youtu.be/EI3SebxlOBs
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/4RsSWWXpGCo
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https://youtu.be/tJm9KqYG4fU?t=3114
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https://youtu.be/JmijOZfwM_A
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 +
~savannahsolver
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https://www.youtube.com/watch?v=2G9jiu5y5PM  ~David
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==See Also==
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{{AMC8 box|year=2017|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 18:20, 15 April 2023

Problem

An integer between $1000$ and $9999$, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

$\textbf{(A) }\frac{14}{75}\qquad\textbf{(B) }\frac{56}{225}\qquad\textbf{(C) }\frac{107}{400}\qquad\textbf{(D) }\frac{7}{25}\qquad\textbf{(E) }\frac{9}{25}$

Solution

There are $5$ options for the last digit as the integer must be odd. The first digit now has $8$ options left (it can't be $0$ or the same as the last digit). The second digit also has $8$ options left (it can't be the same as the first or last digit). Finally, the third digit has $7$ options (it can't be the same as the three digits that are already chosen).

Since there are $9,000$ total integers, our answer is \[\frac{8 \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.\]

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/EI3SebxlOBs

~Education, the Study of Everything

Video Solution

https://youtu.be/4RsSWWXpGCo

https://youtu.be/tJm9KqYG4fU?t=3114

https://youtu.be/JmijOZfwM_A

~savannahsolver

https://www.youtube.com/watch?v=2G9jiu5y5PM ~David

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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