Difference between revisions of "2008 AMC 8 Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | There are <math>2</math> ways to place the remaining <math>\text{ | + | There are <math>2</math> ways to place the remaining <math>\text{As}</math>, <math>2</math> ways to place the remaining <math>\text{Bs}</math>, and <math>1</math> way to place the remaining <math>\text{Cs}</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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:10, 13 November 2017
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?
![[asy] size((80)); draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((0,3)--(9,3)); draw((0,6)--(9,6)); label("A", (1.5,7.5)); [/asy]](http://latex.artofproblemsolving.com/a/8/b/a8bdf881057552dc4d90a565dcc73fef0f5ccf5f.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
