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Difference between revisions of "2015 AMC 12B Problems/Problem 22"

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Label the people sitting at the table <math>A, B, C, D, E, F,</math> and assume that they are initially seated in the order <math>ABCDEF</math>. The possible new positions for <math>A, B, C, D, E,</math> and <math>F</math> are respectively (a dash indicates a non-allowed position):
 
Label the people sitting at the table <math>A, B, C, D, E, F,</math> and assume that they are initially seated in the order <math>ABCDEF</math>. The possible new positions for <math>A, B, C, D, E,</math> and <math>F</math> are respectively (a dash indicates a non-allowed position):
  
−
<math>\texttt{--AAA-}\\
+
<cmath>\begin{tabular}{| c | c | c | c | c | c |}
−
\texttt{---BBB}\\
+
\hline
−
\texttt{C---CC}\\
+
- & - & A & A & A & - \\ \hline
−
\texttt{DD---D}\\
+
- & - & - & B & B & B \\ \hline
−
\texttt{EEE---}\\
+
C & - & - & - & C & C \\ \hline
−
\texttt{-FFF--}</math>
+
D & D & - & - & - & D \\ \hline
 +
E & E & E & - & - & - \\ \hline
 +
- & F & F & F & - & - \\ \hline
 +
\end{tabular}</cmath>
  
 
The permutations we are looking for should use one letter from each column, and there should not be any repeated letters:
 
The permutations we are looking for should use one letter from each column, and there should not be any repeated letters:

Revision as of 22:22, 4 March 2015

Problem

Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?

$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$

Solution

Label the people sitting at the table $A, B, C, D, E, F,$ and assume that they are initially seated in the order $ABCDEF$. The possible new positions for $A, B, C, D, E,$ and $F$ are respectively (a dash indicates a non-allowed position):

\[\begin{tabular}{| c | c | c | c | c | c |} \hline - & - & A & A & A & - \\ \hline - & - & - & B & B & B \\ \hline C & - & - & - & C & C \\ \hline D & D & - & - & - & D \\ \hline E & E & E & - & - & - \\ \hline - & F & F & F & - & - \\ \hline \end{tabular}\]

The permutations we are looking for should use one letter from each column, and there should not be any repeated letters:

CDEFAB
CEAFBD
CEFABD
CEFBAD
CFEABD
CFEBAD
DEAFBC
DEAFCB
DEFABC
DEFACB
DEFBAC
DFEABC
DFEACB
DFEBAC
EDAFBC
EDAFCB
EDFABC
EDFACB
EDFBAC
EFABCD

There are $\boxed{\textbf{(D)}\; 20}$ such permutations.

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 AMC 12 Problems and Solutions

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