Difference between revisions of "2004 AMC 8 Problems/Problem 2"
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== Problem == | == Problem == | ||
| − | How many different four-digit numbers can be formed | + | 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 == | ||
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. | ||
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. | ||
| − | + | <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>. | |
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| + | == 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>. | ||
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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 
?
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 
choices for the first digit, because 
isn't a valid choice, or the number would a 3-digit number, which is not what we want.
We have 
choices for the second digit, since we already used up one of the digits, and choices for the third, and finally just 
choices for the fourth and final one.
is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get 
.
Solution 2
Note that the four-digit number must start with either a or a 
. The four-digit numbers that start with are 
, and . The four-digit numbers that start with are 
, and 
which gives us a total of .
Solution 3
In order for the resulting numbers to have four digits, they cannot start with . Therefore, both zeroes must be in the last three places. There are 
ways to choose which two of the last three places are zeroes. Then there are 
ways to arrange the and the in the remaining two places, giving us a total of 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
