AMC 8 2020 Problem 10
AMC8 2020 年第 10 题
排列与组合★★☆☆☆
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?
Zara 有 4 个玻璃球，分别叫 Aggie，Bumblebee，Steelie 和 Tiger。她想把它们在架子上排成一行，但不想让 Steelie 和 Tiger 相邻。那么她有多少种排列方法？
- A.6
- B.8
- C.12
- D.18
- E.24
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and , respectively. If we ignore the constraint that and cannot be next to each other, we get a total of 4!=24 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 6×2=12 ways to place and next to each other. Subtracting this from the total number of arrangements gives us 24-12=12 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 6×2=12 ways to place and such that they are not next to each other .
Let's try complementary counting. There 4! 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, This solution is best to save time. (edits by Shriyans.Chowdhury)
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 2! ways to arrange Steelie and Tiger within this "super marble." Then there are 3! ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are 4! ways to arrange the marbles without any restrictions, the answer is given by -franzliszt
By the Georgeooga-Harryooga Theorem, our answer is . -franzliszt There are 4!=24 ways to arrange all of the marbles. Since 2 is half of 4, the Steelie and Tiger (2 marbles) are half of the 4 marbles total. So, the ways to arrange them would also be half of the total. Half of 24 is 12, so the answer is 12, or . Let Aggie, Bumblebee, Steelie, and Tiger, be $A,B,S,$ and , respectively. Since and can't be placed next to each other, we first place and with no constraints. We draw __$,A,$__$,B,$__ on the paper. We first choose where goes with 3 choices and then with 2 choices. Since and can also be placed in the other order, the answer is .
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