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 $6$ $\Rightarrow \boxed{\textbf{(B)}\ 6}$

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2004 AMC 8 Problems/Problem 2

Problem

Solution