Difference between revisions of "2017 AMC 8 Problems/Problem 20"
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− | ==Problem | + | ==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? | ||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | There are <math>5</math> options for the last digit | + | 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). |
− | Since there are <math> | + | 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> |
− | ~ | + | ==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== | ||
+ | {{AMC8 box|year=2017|num-b=19|num-a=21}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 19:20, 15 April 2023
Contents
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.