2017 AMC 8 Problems/Problem 20
Contents
[hide]Problem
An integer between and , inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?
Solution
There are options for the last digit as the integer must be odd. The first digit now has options left (it can't be or the same as the last digit). The second digit also has options left (it can't be the same as the first or last digit). Finally, the third digit has options (it can't be the same as the three digits that are already chosen).
Since there are total integers, our answer is
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/tJm9KqYG4fU?t=3114
~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.