Difference between revisions of "2017 AMC 8 Problems/Problem 20"

Latest revision as of 22:11, 13 August 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?

$\textbf{(A) }\frac{14}{75} \qquad \textbf{(B) }\frac{50}{225} \qquad \textbf{(C) }\frac{107}{400} \qquad \textbf{(D) }\frac{7}{25} \qquad \textbf{(E) }\frac{9}{25}$

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-==Problem 20==
+==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?
-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{50}{225} \qquad \textbf{(C) }\frac{107}{400} \qquad \textbf{(D) }\frac{7}{25} \qquad \textbf{(E) }\frac{9}{25}</math>
-<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==
+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>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
-==Solution==
+~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}}
-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).
+{{MAA Notice}}
-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>
+[[Category:Introductory Combinatorics Problems]]
-~nukelauncher
+[[Category:Introductory Probability Problems]]

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