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Suppose we have $M_1, F_1, M_2, F_2, M_3, F_3, M_4, F_4$ to be arranged in a row.

Can this question be solved without inclusion-exclusion e.g. $1-(Couple_1 \cup C_2 \cup C_3 \cup C_4)$

Why is inclusion-exclusion the preferred method for this question?

My attempt: There are 8 choices for the first position, 6 for the next ($7-$the first person's partner), and 5 for the third ($6-$the second person's partner). Then what?

B Green
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    Your method gets messy, fast. Starting $M_1M_2M_3$ leaves you $4$ choices for the next (namely $F_1, F_2, M_4, F_4$). But starting $M_1M_2F_1$ leaves you $5$ choices, (namely $F_2, M_3, F_3, M_4, F_4$). – lulu Jul 06 '21 at 22:13
  • It is $\frac{12}{35}8! = 13824$ as explained in the many questions linked to https://math.stackexchange.com/questions/465318/showing-probability-no-husband-next-to-wife-converges-to-e-1 – Henry Jul 06 '21 at 23:39

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