This problem from American Mathematics Contest 8 (AMC 8, 2018) is based on calculation of probability. It is Question number 11 of the AMC 8 2018 Problem series.
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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?
$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}$
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For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in $4!$ ways which results in a total of $3 \times 2 \times 4!$ ways to arrange them.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]By the same logic, there are 4 ways for Abby and Bridget to placed if they are adjacent in the same row, they can swap seats, and the other $4$ people can be arranged in $4!$ ways, for a total of $4 \times 2 \times 4!$ ways to arrange them.[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]We sum the 2 possibilities up to get
$\frac{(3\cdot2)\cdot4!+(4\cdot2)\cdot4!}{6!} = \frac{14\cdot4!}{6!}=\boxed{\frac{7}{15}}$[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" global_module="50833"][et_pb_fullwidth_header title="AMC - AIME Program" button_one_text="Learn More" button_one_url="https://cheenta.com/amc-aime-usamo-math-olympiad-program/" header_image_url="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.2.2" title_level="h2" background_color="#00457a" custom_button_one="on" button_one_text_color="#44580e" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]AMC - AIME - USAMO Boot Camp for brilliant students. Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
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