2009 AMC 12B Problems/Problem 21

From AoPSWiki

Revision as of 20:45, 27 February 2009 by Azjps (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

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 $W_{10} = \boxed{89}$ .

2009 AMC 12B (Problems • 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

Personal tools

Navigation

Search

Toolbox

2009 AMC 12B Problems/Problem 21

From AoPSWiki

Problem

Solution

See Also