Difference between revisions of "2017 AMC 10A Problems/Problem 19"
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Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of <math>2 \cdot 2 \cdot 2 \cdot 2 = 16</math>. | Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of <math>2 \cdot 2 \cdot 2 \cdot 2 = 16</math>. | ||
| − | <math>\textbf{Case 2: } A does not sit in an edge seat. | + | <math>\textbf{Case 2: }</math> A does not sit in an edge seat. |
| − | In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are < | + | In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are <math>3 \cdot 2 \cdot 2 = 12</math> seatings in this case. |
| − | Adding up all the cases, we have < | + | Adding up all the cases, we have <math>16+12 = \boxed{\textbf{(C) } 28}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:43, 8 February 2017
Problem
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?
Solution
For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E. We can split this problem up into two cases:
A sits on an edge seat.
Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of 
.
A does not sit in an edge seat.
In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are 
seatings in this case.
Adding up all the cases, we have 
.
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
