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| − | ==Problem==
| + | #REDIRECT [[2020 AMC 10B Problems/Problem 17]] |
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| − | There are <math>10</math> people standing equally spaced around a circle. Each person knows exactly <math>3</math> of the other <math>9</math> people: the <math>2</math> people standing next to her or him,as well as the person directly across the circle. How many ways are there for the <math>10</math> people to split up into <math>5</math> pairs so that the members of each pair know each other?
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| − | <math>\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15</math>
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| − | ==Solution==
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| − | We have 2 cases. Assume these people are A,B,...,J.
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| − | Case 1: Group with people nearby: we have (AB)(CD)(EF)(GH)(IJ) and (AJ)(IH)(GF)(ED)(CB), totally 2 cases.
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| − | Case 2: (AF) is a group. So here we have 3 sub-cases:
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| − | Sub 1: The rest of them group with people nearby, that is (BC)(DE)(GH)(IJ), we have 5 cases since (AF) can be (BG), (CH), (DI) and (EJ).
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| − | Sub 2: (BG)(CH) are groups, the rest groups are (DE)(IJ). We also have 5 cases for the same reason of Sub 1.
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| − | Sub 3: (AF)(BG)(CH)(DI)(EJ), 1 case.
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| − | Thus, the total number of cases are 2+5+5+1=13 cases. So put <math>\boxed{C}.</math>
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| − | ~FANYUCHEN20020715
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| − | ==Solution 2==
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| − | Take cases on the number of pairs of people who shake hand across the table.
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| − | Case 1: No pairs. There are two ways.
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| − | Case 2: One pair. There's five ways to choose this pair, and afterwards the remaining 8 people are determined -> 5.
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| − | Case 3: Two pairs. There are no ways to do this.
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| − | Case 4: Three pairs. We see that the three pairs must be adjacent, so there are 5 ways to do this.
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| − | Case 5: Four pairs. Again no ways.
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| − | Case 6: All five pairs. One way.
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| − | Thus, the total number is 2+5+5+1=13 ways, and the anser is <math>\boxed{C}.</math>
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| − | ~ rzlng
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| − | ==See Also==
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| − | {{AMC12 box|year=2020|ab=B|num-b=14|num-a=16}}
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| − | {{MAA Notice}}
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