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

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== Problem ==
 
== Problem ==
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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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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 contains one of each letter. If <math>\text{A}</math> is placed in the upper left corner, how many arrangements are possible?
  
 
<asy>
 
<asy>
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== Solution ==
 
== Solution ==
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There are 2 ways to place the remaining A's 2 ways to place the remaining B's and 1 way to place the remaining C's for a total of (2)(2)(1) =4
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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>.
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==Video Solution==
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https://www.youtube.com/watch?v=8qzMymleTIg  ~David
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==Video Solution 2==
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https://youtu.be/1m_c_iMvxKo  Soo, DRMS, NM
  
 
== 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}}

Latest revision as of 20:28, 1 January 2024

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 contains 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 $(2)(2)(1) = \boxed{\textbf{(C)}\ 4}$.

Video Solution

https://www.youtube.com/watch?v=8qzMymleTIg ~David

Video Solution 2

https://youtu.be/1m_c_iMvxKo Soo, DRMS, NM

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. AMC logo.png

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