Difference between revisions of "2009 AMC 12B Problems/Problem 21"
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<math>\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8</math> | <math>\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8</math> | ||
− | == Solution == | + | == Solution 1== |
− | Let <math>W_n</math> be the answer for <math>n</math> women | + | 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. | ||
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We get the recurrence <math>W_n = W_{n-1} + W_{n-2}</math>. (Hence <math>W_n</math> is precisely the <math>n</math>-th [[Fibonacci number]].) Using this recurrence we can easily compute that <math>W_{10} = \boxed{89}</math>. | We get the recurrence <math>W_n = W_{n-1} + W_{n-2}</math>. (Hence <math>W_n</math> is precisely the <math>n</math>-th [[Fibonacci number]].) Using this recurrence we can easily compute that <math>W_{10} = \boxed{89}</math>. | ||
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+ | ==Solution 2== | ||
+ | Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done by recursion as in Solution 1, or we can casework depending on the number of 2x1 blocks. The cases of 1, 2, 3, 4, 5 2x1 blocks correspond to 8, 6, 4, 2, 0 1x1 blocks, and so the answer is | ||
+ | <cmath>\binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 9 + 28 + 35 + 15 + 1 = 89.</cmath> | ||
== See Also == | == See Also == |
Revision as of 23:28, 31 January 2015
Contents
Problem
Ten women sit in seats in a line. All of the get up and then reseat themselves using all 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?
Solution 1
Let be the answer for women. We want to find .
Clearly . Now let . Let the row of seats go from left to right. Label both the seats and the women to , going from left to right. Consider the rightmost seat. Which women can sit there after the swap? It can either be woman or woman , as for any other woman the seat is too far.
If woman stays in her seat, there are exactly valid arrangements of the other women. If woman sits on seat , we only have one option for woman : she must take seat , all the other seats are too far for her. We are left with women to sitting on seats to , and there are clearly valid arrangements of these.
We get the recurrence . (Hence is precisely the -th Fibonacci number.) Using this recurrence we can easily compute that .
Solution 2
Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done by recursion as in Solution 1, or we can casework depending on the number of 2x1 blocks. The cases of 1, 2, 3, 4, 5 2x1 blocks correspond to 8, 6, 4, 2, 0 1x1 blocks, and so the answer is
See Also
2009 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 | |
All AMC 12 Problems and Solutions |
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