Difference between revisions of "2008 AMC 8 Problems/Problem 14"
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</asy> | </asy> | ||
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math> | <math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | There are <math>2</math> ways to place the remaining <math>\text{A's}</math>, <math>2</math> ways to place the remaining <math>\text{B's}</math>, and <math>1</math> way to place the remaining <math>\text{C's}</math> for a total of <math>(2)(2)(1) = \boxed{\textbf{(C)}\ 4}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=13|num-a=15}} | {{AMC8 box|year=2008|num-b=13|num-a=15}} |
Revision as of 02:38, 25 December 2012
Problem
Three , three , and three are placed in the nine spaces so that each row and column contain one of each letter. If is placed in the upper left corner, how many arrangements are possible?
Solution
There are ways to place the remaining , ways to place the remaining , and way to place the remaining for a total of .
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |