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=\boxed{\textbf{(D) }18}$
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) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
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| All AJHSME/AMC 8 Problems and Solutions | ||
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