Difference between revisions of "2018 AMC 8 Problems/Problem 11"
(→Solution 2) |
(→Solution) |
||
Line 38: | Line 38: | ||
− | We sum the 2 possibilities up to get <math>\frac{(3 | + | We sum the 2 possibilities up to get <math>\frac{(3\cdot2)\cdot4!+(4\cdot2)\cdot4!}{6!} = \frac{14\cdot4!}{6!}=\boxed{\frac{7}{15}}</math> or <math>\textbf{(C)}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 14:46, 15 July 2019
Problem 11
Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?
Solution
There are a total of ways to arrange the kids.
Abby and Bridget can sit in 3 ways if they are adjacent in the same column:
For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in ways which results in a total of ways to arrange them.
By the same logic, there are 4 ways for Abby and Bridget to placed if they are adjacent in the same row, they can swap seats, and the other people can be arranged in ways, for a total of ways to arrange them.
We sum the 2 possibilities up to get or .
Solution 2
We can ignore about the 4 other classmates because they aren't relevant. We can treat Abby and Bridget as a pair, so there are total ways to seat them. If they sit in the same row, there are ways to seat them. If they sit in the same column, there are ways to seat them. Thus our answer is
Solution 3
Total number of ways n(S) = C(6,2) = 15, if we treat Abby and Bridget as a pair and forget the others. Total number of ways they are adjacent = 4 (for the rows) + 3 (for the columns) Therefore, P(Abby and Bridget sitting adjacent) is 7/15 (C)
See also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.