Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column? $X\ X\ X$
$X\ X\ X$

A. $\frac{1}{3}$

B. $\frac{2}{5}$

C. $\frac{7}{15}$

D. $\frac{1}{2}$

E. $\frac{2}{3}$

1. 2018 AMC 8 官方解析

method 1:

There are $6!$ ways to arrange all students. Abby and Bridget are adjacent to each other in the same column in these three arrangements:

$A\ X\ X\quad X\ A\ X\quad X\ X\ A$
$B\ X\ X\quad X\ B\ X\quad X\ X\ B$

Observe that Abby and Bridget can exchange seats without violating this condition, and that there are $4!$ ways to arrange the other students. Therefore, there are $3 \times 2 \times 4!$ ways for Abby and Bridget to be adjacent in the same column.

Similarly, there are four arrangements in which Abby and Bridget are adjacent to each other in the same row. They can exchange seats without violating this condition, and there are $4!$ ways to arrange the other students. Therefore, there are $4 \times 2 \times 4!$ ways for Abby and Bridget to be adjacent in the same row.Then probability is then,
$\frac{(3 \times 2) \times 4! + (4 \times 2) \times 4!}{6!} = \frac{14 \times 4!}{6!} = \frac{7}{15}$.The correct answer is C.

method 2:

In this method, we practically disregard the other students in our calculation. Let Abby and Bridget "stick together." There are $\binom{6}{2} = 15$ ways to get them seated. Moreover, there are $2 \times 2 = 4$ ways for them to be seated in the same row and three ways in the same column.
The probability is then, $\frac{4+3}{15} = \frac{7}{15}$.The correct answer is C.