2017 AMC 8

Complete problem set with solutions and individual problem pages

Problem 20 Hard

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?

  • A.

    \frac{14}{75}

  • B.

    \frac{56}{225}

  • C.

    \frac{107}{400}

  • D.

    \frac{7}{25}

  • E.

    \frac{9}{25}

Answer:B

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}}.

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