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

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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>
 
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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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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https://youtu.be/EI3SebxlOBs
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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~savannahsolver
 
~savannahsolver
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https://www.youtube.com/watch?v=2G9jiu5y5PM  ~David
  
 
==See Also==
 
==See Also==

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

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