# Difference between revisions of "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 1

We can solve this problem easily, just by calculating how many choices there are for each of the four digits. First off, we know there are only $2$ choices for the first digit, because $0$ isn't a valid choice, or the number would a 3-digit number, which is not what we want. We have $3$ choices for the second digit, since we already used up one of the digits, and $2$ choices for the third, and finally just $1$ choices for the fourth and final one. Now we all $2+3+2+1$, which is $\boxed{\textbf{(B)}\ 6}$.

## Solution 2

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 byProblem 1 Followed byProblem 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

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