Difference between revisions of "2020 AMC 8 Problems/Problem 10"
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− | Back to the problem. | + | Back to the problem. Our answer is <math>\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\textbf{(C) }12</math>. |
-franzliszt | -franzliszt |
Revision as of 18:07, 23 November 2020
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
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 by RedFireTruck
Solution 2
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
Solution 3
Let's try complementary counting. There ways to arrange the 4 marbles. However, there are
arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus,
Solution 4
We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are ways to arrange Steelie and Tiger within this "super marble." Then there are
ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are
ways to arrange the marbles without any restrictions, the answer is given by
-franzliszt
Solution 5
We will use the following
The Georgeooga-Harryooga Theorem states that if you have
distinguishable objects and
of them cannot be together, then there are
ways to arrange the objects.
Let our group of objects be represented like so
,
,
, ...,
,
. Let the last
objects be the ones we can't have together.
Then we can organize our objects like so .
We have ways to arrange the objects in that list.
Now we have blanks and
other objects so we have
ways to arrange the objects we can't put together.
By fundamental counting principal our answer is .
Proof by RedFireTruck
Back to the problem. Our answer is .
-franzliszt
Video Solution
~savannahsolver
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 |
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