2020 AMC 8 Problems/Problem 10
Contents
Problem
Zara has a collection of marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
Solution 1
First we need to count the the total number of possibilities. That is 4!=24. Now we subtract the cases where the Tiger and the Steelie are together. Doing this, we get 24-6=18 or D.
Solution 2 (complementary counting)
There would be ways to arrange the marbles, except for the condition that the Steelie and Tiger cannot be next to each other. If we did place them next to each other with the Steelie first, there would be ways to place them (namely , ; or , where and denote the Steelie and the Tiger as in Solution 1). Accounting for the other possible order, there are a total of ways. Now, there are ways to place and , giving overall ways to arrange the marbles with and next to each other. Subtracting this from (to remove the cases which are not allowed) gives valid ways to arrange the marbles.
Solution 3 (variant of Solution 2)
As in Solution 2, there are total ways to arrange the marbles without any constraints. To count the number of ways where the Steelie and the Tiger are next to each other, we treat them together as a "super marble". There are ways to arrange the Steelie and Tiger within the super marble, then ways to put the super marble in a row with the Aggie and the Bumblebee. Thus the answer is .
Video Solution
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
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Followed by Problem 11 | |
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