Bob and Alice are dinner guests at a party of eight, 4 male and 4 female. The hostess arranges the guests linearly along a table with the men on one side and the women on the other. The probability that Bob and Alice will be facing each other or be within one position of facing each other is...

  • $\begingroup$ Hard to say. I would imagine not all possible arrangements are equally likely, as there's a good chance at least one couple is coming, and will be seated together. $\endgroup$ – TBrendle Sep 29 '13 at 18:45

There are $6$ forbidden places places that Alice and Bob could sit, illustrated below:

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Once they are seated, the remaining people can be seated in $3!^2$ distinct ways.

The total number of seating arrangements (forbidden or not) is $4!^2$.

Hence the probability of a permissible seating arrangement is therefore $$1-\frac{6 \times 3!^2}{4!^2}=1-\frac{216}{576}=0.625.$$

(Note: I've assumed that there is a designated male side and a female side here; if the restriction is only that "the males are on one side", then both the numerator and denominator above should be multiplied by $2$.)

  • $\begingroup$ Hello, thanks for the help. But I think the question is asking that Bob and Alice must either be sitting across from each other or at least one seat adjacent from each other. $\endgroup$ – user2827137 Sep 29 '13 at 19:01
  • $\begingroup$ Oops! I've done the opposite; well the probability will be $1-0.375$. It should be fixed now. $\endgroup$ – Rebecca J. Stones Sep 29 '13 at 19:04
  • $\begingroup$ Your missing a square in your equation, 4!. $\endgroup$ – wtsang02 Apr 4 '14 at 17:10

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