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

(Solution 2)
(Solution 2)
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== Solution  1==
 
== 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>.
== Solution 2 ==
+
== Solution 1 ==
 
We can solve this problem easily, just by calculating how many choices there are for each of the four digits.
 
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 <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.  

Revision as of 04:26, 24 July 2018

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

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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