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"?
$\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}$
Solution 1
There is one favorable case, which is the license plate says "AMC8."  We must now find how many total cases there are.  There are $5$ choices for the first letter (since it must be a vowel), $21$ choices for the second letter (since it must be of $21$ consonants), $20$ choices for the third letter (since it must differ from the second letter), and $10$ choices for the digit.  This leads to $5 \cdot 21 \cdot 20 \cdot 10=21,000$ total possible license plates.  Therefore, the probability of a license plate saying "AMC8" is $\boxed{\textbf{(B) } \frac{1}{21,000}}$.

Solution 2
The probability of choosing A as the first letter is $\dfrac{1}{5}$.  The probability of choosing $M$ next is $\dfrac{1}{21}$.  The probability of choosing C as the third letter is $\dfrac{1}{20}$ (since there are $20$ other consonants to choose from other than M). The probability of having $8$ as the last number is $\dfrac{1}{10}$.  We multiply all these to obtain $\dfrac{1}{5}\cdot \dfrac{1}{21} \cdot \dfrac{1}{20}\cdot \dfrac{1}{10}=\dfrac{1}{5\times 21\times 20\times 10}=\dfrac{1}{21\times 100\times 10}=\boxed{\textbf{(B)}~\dfrac{1}{21,000}}$.