2004 AMC 8 Problems/Problem 2

Problem

How many different four-digit numbers can be formed be rearranging the four digits in $2004$?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$

Solution

Note that the four-digit number must start with either a $2$ or a $4$. The four-digit numbers that start with $2$ are $2400, 2040$, and $2004$. The four-digit numbers that start with $4$ are $4200, 4020$, and $4002$ which gives us a total of $\boxed{\textbf{(B)}\ 6}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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