Difference between revisions of "2017 AMC 8 Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | 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). | + | 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>9000</math> total integers, out 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, out answer is <cmath>\frac{8 \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.</cmath> | ||
Revision as of 21:55, 22 November 2017
Problem 20
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, out answer is ![\[\frac{8 \cdot 8 \cdot 7 \cdot 5}{9000} = \boxed{\textbf{(B)}\ \frac{56}{225}}.\]](http://latex.artofproblemsolving.com/5/d/8/5d8a4f26914ee2febf86c13a27c746c91b00f952.png)
~nukelauncher
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. 
