# 2017 AMC 8 Problems/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? $\textbf{(A) }\frac{14}{75}\qquad\textbf{(B) }\frac{56}{225}\qquad\textbf{(C) }\frac{107}{400}\qquad\textbf{(D) }\frac{7}{25}\qquad\textbf{(E) }\frac{9}{25}$

## 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!!!)

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## Video Solution

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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 