AMC8 2020

AMC8 2020 · Q10

AMC8 2020 · Q10. It mainly tests Basic counting (rules of product/sum), Inclusion–exclusion (basic).

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 颗弹珠：一颗 Aggie、一颗 Bumblebee、一颗 Steelie 和一颗 Tiger。她想把它们排成一排放在架子上，但不想让 Steelie 和 Tiger 紧挨着。有多少种方式可以做到这一点？

(A) 6 6

(B) 8 8

(D) 18 18

(E) 24 24

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$, respectively. If we ignore the constraint that $S$ and $T$ 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 $S$ and $T$ can be next to each other. If we place $S$ and $T$ 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. $ST\square\square, \square ST\square, \square\square ST$). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$. Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\implies\boxed{\textbf{(C) }12}$. We can also solve this problem directly by looking at the number of ways that we can place $S$ and $T$ such that they are not directly next to each other. Observe that there are three ways to place $S$ and $T$ (in that order) into the four slots so they are not next to each other (i.e. $S\square T\square, \square S\square T, S\square\square T$). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $T\square S\square, \square T\square S, T\square\square S$). Thus there are 6 ways of placing $S$ and $T$ so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$. Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ such that they are not next to each other $\implies\boxed{\textbf{(C) }12}$.

设 Aggie、Bumblebee、Steelie 和 Tiger 分别称为 $A,B,S,$ 和 $T$。如果忽略 $S$ 和 $T$ 不能紧挨着的约束，总共有 $4!=24$ 种方式排列 4 颗弹珠。现在只需减去 $S$ 和 $T$ 紧挨着的方式数。如果将 $S$ 和 $T$ 按此顺序放在一起，有三个位置：前两个槽、第二和第三个槽，或最后两个槽（即 $ST\square\square, \square ST\square, \square\square ST$）。但也可以反序 $TS\square\square, \square TS\square, \square\square TS$。因此有 6 种放置 $S$ 和 $T$ 紧挨着的方式。对于每种放置，剩下两个槽放置 $A$ 和 $B$ 有 $2!$ 种方式，即 $6\times 2=12$ 种紧挨方式。总排列减去此数得 $24-12=12$，即 $\boxed{\textbf{(C) }12}$。也可以直接计算 $S$ 和 $T$ 不紧挨的方式。有三种方式将 $S$ 和 $T$（按此序）放入四槽不紧挨（即 $S\square T\square, \square S\square T, S\square\square T$），反序也有三种，共 6 种。然后每个有 $2!$ 种填 $A,B$，总 $6\times 2=12$，即 $\boxed{\textbf{(C) }12}$。

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