Difference between revisions of "2009 AMC 12B Problems/Problem 21"

Revision as of 08:43, 27 February 2009

Problem

Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?

$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$

Solution

Let $W_n$ be the answer for $n$ women, we want to find $W_{10}$ .

Clearly $W_0=W_1=1$ . Now let $n>1$ . Let the row of seats go from left to right. Label both the seats and the women $1$ to $n$ , going from left to right. Consider the rightmost seat. Which women can sit there after the swap? It can either be woman $n$ or woman $n-1$ , as for any other woman the seat is too far.

If woman $n$ stays in her seat, there are exactly $W_{n-1}$ valid arrangements of the other $n-1$ women. If woman $n-1$ sits on seat $n$ , we only have one option for woman $n$ : she must take seat $n-1$ , all the other seats are too far for her. We are left with women $1$ to $n-2$ sitting on seats $1$ to $n-2$ , and there are clearly $W_{n-2}$ valid arrangements of these.

We get the recurrence $W_n = W_{n-1} + W_{n-2}$ . (Hence $W_n$ is precisely the $n$ -th Fibonacci number.) Using this recurrence we can easily compute that $W_{10} = \boxed{89}$ .

@@ Line 8: / Line 8: @@
 Let <math>W_n</math> be the answer for <math>n</math> women, we want to find <math>W_{10}</math>.
-Clearly <math>W_0=W_1=1</math>. Now let <math>n>1</math>. Let the row of seats go from left to right. Label both the seats and the women <math>1</math> to <math>N</math>, going from left to right. Consider the rightmost seat. Which women can sit there after the swap? It can either be woman <math>n</math> or woman <math>n-1</math>, as for any other woman the seat is too far.
+Clearly <math>W_0=W_1=1</math>. Now let <math>n>1</math>. Let the row of seats go from left to right. Label both the seats and the women <math>1</math> to <math>n</math>, going from left to right. Consider the rightmost seat. Which women can sit there after the swap? It can either be woman <math>n</math> or woman <math>n-1</math>, as for any other woman the seat is too far.
 If woman <math>n</math> stays in her seat, there are exactly <math>W_{n-1}</math> valid arrangements of the other <math>n-1</math> women. If woman <math>n-1</math> sits on seat <math>n</math>, we only have one option for woman <math>n</math>: she must take seat <math>n-1</math>, all the other seats are too far for her. We are left with women <math>1</math> to <math>n-2</math> sitting on seats <math>1</math> to <math>n-2</math>, and there are clearly <math>W_{n-2}</math> valid arrangements of these.

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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
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Difference between revisions of "2009 AMC 12B Problems/Problem 21"

Revision as of 08:43, 27 February 2009

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