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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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Since there are <math>9000</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>9000</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==
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https://youtu.be/tJm9KqYG4fU?t=3114
  
 
==See Also==
 
==See Also==

Revision as of 14:10, 18 January 2021

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

Video Solution

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

See Also

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

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