Difference between revisions of "2017 AMC 8 Problems/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?
 
 
<math>\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}</math>
 
 
 
==Solution==
 
==Solution==
  

Latest revision as of 20:50, 12 December 2024

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 (ProblemsAnswer KeyResources)
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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