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

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

Solution 2

There is only 2 choices for the first digit because you can't have 0 as the first digit because it wouldn't be a 4 digit number. Then there are 3 choices for the second and 2 for the third and 1 for the fourth. Then just like you would do for how many ways there are to arrange a word with two of the same letters, you do $\frac{2 \cdot 3 \cdot 2 \cdot1}{2!}$ . Which is $\frac{12}{2}$ which is simplified to $\boxed{\textbf{(B)}\ 6}$

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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2004 AMC 8 Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See Also