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 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 == | ||
| + | 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 18:49, 10 November 2016
Contents
Problem
How many different four-digit numbers can be formed be rearranging the four digits in 
?
Solution 1
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 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 
. Which is 
which is simplified to
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. 
