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

Write $S$ and $T$ for the Steelie and the Tiger respectively. Putting in $S$ and $T$ first, in order to avoid them being next to each other, we must have the arrangement $S\square T\square$ , $S\square\square T$ , $\square S\square T$ , or any of these with the $S$ and $T$ swapped. This gives $3 \cdot 2 = 6$ ways, and there are then $2$ ways to put in the Aggie and Bumblebee, for a total of $2\cdot 6=\boxed{\textbf{(C) }12}$ .

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 $24-2\cdot 6=\boxed{\textbf{(C) }12}$ .

Solution 4(Using Probability)

To solve this problem, we can first consider the chance of this happening and try to involve it in the total possibilities. As we know, we do not want the Steelie and the Tiger together, the total outcomes are that they are together and that they are not together. As seen in the previous solutions, them being together and not together total $24$ possibilities with them both being $12$ separate ways. To solve this problem, we must consider the possibility of this happening and multiply it by the total ways to get an accurate result. The probability of this happening would be $\frac{1}{2}$ since they can either be together or not. This is also seen in Pascal's Triangle as the sum of the numbers in each row result in powers of $2$ , where each entity can be a part of the total or not(for committees) but this also applies to general counting as well. The total number of ways without restrictions is $4!=24$ , and by multiplying it by $\frac{1}{2}$ , we get $24 \cdot \frac{1}{2}=12$ .

Video Solution

https://youtu.be/61c1MR9tne8

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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

Page

Toolbox

Search