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Difference between revisions of "2004 AMC 8 Problems/Problem 2"

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== Problem ==
 
== Problem ==
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How many different four-digit numbers can be formed be rearranging the four digits in <math>2004</math>?
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How many different four-digit numbers can be formed by rearranging the four digits in <math>2004</math>?
  
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81 </math>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81 </math>
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== Solution 1 ==
 
== Solution 1 ==
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First off, we know there are only <math>2</math> choices for the first digit, because <math>0</math> isn't a valid choice, or the number would a 3-digit number, which is not what we want.  
 
First off, we know there are only <math>2</math> choices for the first digit, because <math>0</math> isn't a valid choice, or the number would a 3-digit number, which is not what we want.  
 
We have <math>3</math> choices for the second digit, since we already used up one of the digits, and <math>2</math> choices for the third, and finally just <math>1</math> choices for the fourth and final one.  
 
We have <math>3</math> choices for the second digit, since we already used up one of the digits, and <math>2</math> choices for the third, and finally just <math>1</math> choices for the fourth and final one.  
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Now we all <math>2+3+2+1</math>, which is <math>\boxed{\textbf{(B)}\ 6}</math>.
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<math>2*3*2*1</math> is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get <math>\boxed{\textbf{(B)}\ 6}</math>.
  
 
== Solution  2==
 
== Solution  2==
 
Note that the four-digit number must start with either a <math>2</math> or a <math>4</math>. The four-digit numbers that start with <math>2</math> are <math>2400, 2040</math>, and <math>2004</math>. The four-digit numbers that start with <math>4</math> are <math>4200, 4020</math>, and <math>4002</math> which gives us a total of <math>\boxed{\textbf{(B)}\ 6}</math>.
 
Note that the four-digit number must start with either a <math>2</math> or a <math>4</math>. The four-digit numbers that start with <math>2</math> are <math>2400, 2040</math>, and <math>2004</math>. The four-digit numbers that start with <math>4</math> are <math>4200, 4020</math>, and <math>4002</math> which gives us a total of <math>\boxed{\textbf{(B)}\ 6}</math>.
  
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==Solution 3==
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 +
In order for the resulting numbers to have four digits, they cannot start with <math>0</math>. Therefore, both zeroes must be in the last three places. There are <math>\binom32=3</math> ways to choose which two of the last three places are zeroes. Then there are <math>2\cdot1=2</math> ways to arrange the <math>2</math> and the <math>4</math> in the remaining two places, giving us a total of <math>3\cdot2=\boxed{\textbf{(B)}\ 6}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|num-b=1|num-a=3}}
 
{{AMC8 box|year=2004|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:16, 21 July 2024

Problem

How many different four-digit numbers can be formed by 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. $2*3*2*1$ is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get $\boxed{\textbf{(B)}\ 6}$.

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

Solution 3

In order for the resulting numbers to have four digits, they cannot start with $0$. Therefore, both zeroes must be in the last three places. There are $\binom32=3$ ways to choose which two of the last three places are zeroes. Then there are $2\cdot1=2$ ways to arrange the $2$ and the $4$ in the remaining two places, giving us a total of $3\cdot2=\boxed{\textbf{(B)}\ 6}$.

See Also

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

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