2017 AMC 8 Problems/Problem 20: Difference between revisions
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==Problem== | ==Problem== | ||
An integer between 1000 and 9999, inclusive, is chosen at random. What is the probability that it | 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? <math>(A) 14 | 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== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Combinatorics Problems]] | |||
[[Category:Introductory Probability Problems]] | |||
Latest revision as of 23:49, 12 September 2025
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?
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 | ||
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