Difference between revisions of "2020 AMC 8 Problems/Problem 10"
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<math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math> | <math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math> | ||
==Solution 1== | ==Solution 1== | ||
| − | By the [[Georgeooga-Harryooga Theorem]] there are <math>\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}</math> way to arrange the | + | By the [[Georgeooga-Harryooga Theorem]] there are <math>\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}</math> way to arrange the marbles. |
| + | |||
==Solution 2== | ==Solution 2== | ||
We can arrange our marbles like so <math>\square A\square B\square</math>. | We can arrange our marbles like so <math>\square A\square B\square</math>. | ||
Revision as of 09:45, 18 November 2020
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?
Contents
Solution 1
By the Georgeooga-Harryooga Theorem there are 
way to arrange the marbles.
Solution 2
We can arrange our marbles like so 
.
To arrange the 
and 
we have 
ways.
To place the 
and 
in the blanks we have 
ways.
By fundamental counting principle our final answer is 
Solution 3
Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by 
and , respectively. If we ignore the constraint that and cannot be next to each other, we get a total of 
ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that and can be next to each other. If we place and next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. 
). However, we could also have placed and in the opposite order (i.e. 
). Thus there are 6 ways of placing and directly next to each other. Next, notice that for each of these placements, we have two open slots for placing and . Specifically, we can place in the first open slot and in the second open slot or switch their order and place in the first open slot and in the second open slot. This gives us a total of 
ways to place and next to each other. Subtracting this from the total number of arrangements gives us 
total arrangements 
.
We can also solve this problem directly by looking at the number of ways that we can place and such that they are not directly next to each other. Observe that there are three ways to place and (in that order) into the four slots so they are not next to each other (i.e. 
). However, we could also have placed and in the opposite order (i.e. 
). Thus there are 6 ways of placing and so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing and . Specifically, we can place in the first open slot and in the second open slot or switch their order and place in the first open slot and in the second open slot. This gives us a total of ways to place and such that they are not next to each other .
~ junaidmansuri
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
