Difference between revisions of "2009 AMC 12B Problems/Problem 21"
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== Problem == | == Problem == | ||
− | Ten women sit in <math>10</math> seats in a line. All of the <math>10</math> get up and then reseat themselves using all <math>10</math> 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? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Ten women sit in <math>10</math> seats in a line. All of the <math>10</math> get up and then reseat themselves using all <math>10</math> 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?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<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== |
+ | 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> | ||
− | |||
− | + | ==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> | ||
− | + | Starting with <math>S_1=1</math> and <math>S_2=2,</math> we can calculate <math>S_{10}=\boxed{89}.</math> | |
− | + | ==Clarification of Solution 2== | |
+ | The seating possibilities of woman #<math>10</math> become the two cases which we work out. <math>S_n</math> was defined to be the number of different seating arrangements with <math>n</math> women. | ||
+ | In the first "case" if woman #<math>10</math> sits in the seat #<math>10</math>, this leads to a similar scenario, but with <math>9</math> women instead. That means that for this case, there are a total of <math>S_{9}</math> possible arrangements. We don't know how many exactly, but we are able to define it in terms of <math>S</math>. | ||
+ | |||
+ | During the second "case", woman #<math>10</math> sits in seat #<math>9</math>. This time, woman #9 must go to seat #<math>10</math>, as she is the only other person who can go there. This leaves us with <math>8</math> women, and we again represent this in terms of <math>S \Rightarrow S_8</math>. | ||
+ | |||
+ | Therefore, we can write <math>S_{10}</math> in terms of <math>S_8</math> and <math>S_9</math>, like so: | ||
+ | |||
+ | <cmath>S_{10} = S_8 + S_9.</cmath> | ||
+ | |||
+ | We can then generalize this to say | ||
+ | |||
+ | <cmath>S_n = S_{n-1}+S_{n-2}.</cmath> | ||
+ | |||
+ | Calculating <math>S_1 = 1</math> and <math>S_2 =2,</math> then following the recursive rule from above, we get <math>S_{10} = 89 \Rightarrow \boxed{\text{A}}</math>. | ||
== See Also == | == See Also == | ||
− | {{AMC12 box|year= | + | {{AMC12 box|year=2009|ab=B|num-b=20|num-a=22}} |
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:28, 12 December 2020
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
Clarification of Solution 2
The seating possibilities of woman # become the two cases which we work out. was defined to be the number of different seating arrangements with women.
In the first "case" if woman # sits in the seat #, this leads to a similar scenario, but with women instead. That means that for this case, there are a total of possible arrangements. We don't know how many exactly, but we are able to define it in terms of .
During the second "case", woman # sits in seat #. This time, woman #9 must go to seat #, as she is the only other person who can go there. This leaves us with women, and we again represent this in terms of .
Therefore, we can write in terms of and , like so:
We can then generalize this to say
Calculating and then following the recursive rule from above, we get .
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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