Question

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?

$\dfrac{1}{22,050}$

$\dfrac{1}{21,000}$

$\dfrac{1}{10,500}$

$\dfrac{1}{2,100}$

$\dfrac{1}{1,050}$

Video solution:

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Written solution:

There are $5$ choices for the first symbol, $21$ choices for the second, $20$ choices for the third because it must be a different non-vowel, and $10$ choices for the final digit.

Thus there are $5\cdot21\cdot20\cdot10=21000$ possible plates. Exactly one of these is AMC8, so the probability is $\dfrac{1}{21000}$ .

Thus, B is the correct answer.

2015 AMC 8 Problem 11

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Written solution:

Problem 11 in Other Years