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

(Solution)
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<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>
  
== Solution ==
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== Solution 1==
 
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 2 ==
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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 <math>\frac{2 \cdot 3 \cdot 2 \cdot1}{2!}</math>. Which is <math>\frac{12}{2}</math> which is simplified to <math>\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}}

Revision as of 19:49, 10 November 2016

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

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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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