Difference between revisions of "2004 AMC 8 Problems/Problem 2"
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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 | + | == 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 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. |
Revision as of 04:26, 24 July 2018
Contents
Problem
How many different four-digit numbers can be formed be rearranging the four digits in ?
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 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. Now we all , which is .
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.