Difference between revisions of "2009 AMC 12B Problems/Problem 21"
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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 via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is | 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 via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is | ||
<cmath>\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.</cmath> | <cmath>\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.</cmath> | ||
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+ | ==Solution 2== | ||
+ | Let <math>S_n</math> be the number of possible seating arrangements with <math>n</math> women. Consider <math>n \ge 3,</math> and focus on the rightmost woman. If she returns back to her seat, then there are <math>S_{n-1}</math> ways to seat the remaining <math>n-1</math> women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us <math>S_{n-2}</math> ways to seat the other <math>n-2</math> women, so we obtain the recursion | ||
+ | <cmath>S_n = S_{n-1}+S_{n-2}.</cmath> | ||
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+ | Starting with <math>S_1=1</math> and <math>S_2=2,</math> we can calculate <math>S_{10}=\boxed{89}.</math> | ||
== See Also == | == See Also == |
Revision as of 18:58, 29 January 2016
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
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 via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is
Solution 2
Let be the number of possible seating arrangements with women. Consider and focus on the rightmost woman. If she returns back to her seat, then there are ways to seat the remaining women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us ways to seat the other women, so we obtain the recursion
Starting with and we can calculate
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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