Difference between revisions of "2008 AMC 8 Problems/Problem 14"

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==Problem==
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== Problem ==
 
Three <math>\text{A's}</math>, three <math>\text{B's}</math>, and three <math>\text{C's}</math> are placed in the nine spaces so that each row and column contain one of each letter. If <math>\text{A}</math> is placed in the upper left corner, how many arrangements are possible?
 
Three <math>\text{A's}</math>, three <math>\text{B's}</math>, and three <math>\text{C's}</math> are placed in the nine spaces so that each row and column contain one of each letter. If <math>\text{A}</math> is placed in the upper left corner, how many arrangements are possible?
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<asy>
 
<asy>
 
size((80));
 
size((80));
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label("A", (1.5,7.5));
 
label("A", (1.5,7.5));
 
</asy>
 
</asy>
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<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==
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== Solution ==
 
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>.
 
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==
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== 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}}

Latest revision as of 12:37, 19 October 2020

Problem

Three $\text{A's}$, three $\text{B's}$, and three $\text{C's}$ are placed in the nine spaces so that each row and column contain one of each letter. If $\text{A}$ 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]

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution

There are $2$ ways to place the remaining $\text{As}$, $2$ ways to place the remaining $\text{Bs}$, and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{\textbf{(C)}\ 4}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
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

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