求助一道kaplan的模考题

小小照妖镜

How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?
A. 2
B. 3
C. 4
D. 6
E. 12

给的答案是D，解释：
A common mistake is to pick (A). It's a logic task that many people prefer to visualize, using dashes for desks. We'll just refer to the desks as 1, 2, 3, and 4, in that order, and we'll call the students X and Y. In order to keep at least one empty desk between the two, we have to keep either one or two empty desks between them. If we keep only one desk between the two, there are two ways the chairs can be occupied: X can be in 1 with Y in 3, Y can be in 1 with X in 3, or X can be in 2 with Y in 4, or Y can be 2 with X in 4. If, on the other hand, we keep two empty chairs between the students, then there are two ways to occupy the desks: X can be in 1 withY in 4, or Y can be in 1 with X in 4. So there are 6 different possible arrangements

但是我觉得答案是8啊。
X和Y还可以在2,3，这样他们旁边也各有一个空位啊，对不？

小小照妖镜

= =
啊...真不好意思...
看题不仔细

Kb24

因为问的是there is always at least one empty desk between the students，就是说必须得两个students中间有空位。也就是说两个student不能坐在一起。也可以用减法来做这道题：
总的坐法是4x3=12种。然后要去掉两个student并排坐的情况：3x2=6（左右互换所以x2）
最后得出12-6=6
LZ说的XY在2,3位不符合题目要求，因为两人中间没有空位了

阿抠

one empty desk between the students。“between”