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

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2004 AMC 8 Problems/Problem 2

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