# Difference between revisions of "2020 AMC 8 Problems/Problem 10"

## Problem

Zara has a collection of $4$ 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?

$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

## 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 Stellie are together. Doing this, we get 24-6=$2\cdot 6=\boxed{\textbf{(D) }18}$

## Solution 2 (complementary counting)

There would be $4!=24$ ways to arrange the $4$ 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 $3$ ways to place them (namely $ST\square\square$, $\square ST\square$; or $\square\square ST$, where $S$ and $T$ denote the Steelie and the Tiger as in Solution 1). Accounting for the other possible order, there are a total of $3 \cdot 2 = 6$ ways. Now, there are $2$ ways to place $A$ and $B$, giving overall $6 \cdot 2 = 12$ ways to arrange the marbles with $S$ and $T$ next to each other. Subtracting this from $24$ (to remove the cases which are not allowed) gives $24-12=\boxed{\textbf{(C) }12}$ valid ways to arrange the marbles.

## Solution 3 (variant of Solution 2)

As in Solution 2, there are $24$ 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 $2$ ways to arrange the Steelie and Tiger within the super marble, then $3! = 6$ ways to put the super marble in a row with the Aggie and the Bumblebee. Thus the answer is $24-2\cdot 6=\boxed{\textbf{(C) }12}$.