The theater is configured so that in any row there is a wall, 6 seats, an aisle, 6 more seats, and then a wall. In total there are 2* 6! arrangements = 1440.
"How many ways can they sit if Jane and Mary refuse to sit next to each other?" I have tried to approach this in 3 different ways. First that the number of arrangments is the number of total arrangements w/o restriction minus the configurations of the restriction such that 6! - 5! = 1200 potential arrangements. But this is the difference in the outcomes so I'm not sure how this would represent the separate seating, I think it would represent the together seating.
I have also approached by drawing a grid to represent Mary and Janes seating. mary and jane seating This chart indicates where the two can sit without sitting next to each other, and these outcomes can be multiplied by 2 to account for both sides of the rows. But here I don't understand how to translate the chart into possible outcomes. If we block Mary and Jane together than the amount of outcomes in a row is 5! * 2! which is 240 and then if we multiply by 2 for the row on the other side of the aisle, we get 480 results.
I think this may be the correct solution but I am not sure. If we consider the entire outcomes of all seating arrangements to include the subsets of Mary and Jane sitting together and Mary and Jane sitting together then we can subtract the seated together outcome and end up with the seated apart outcome, and then multiply by 2 to indicate both rows on either side of the aisle. So then the result would be 6! - ( 5! * 2!) = 480 and 480 * 2rows = 960.
Please advise on which logic is correct it would also be helpful to know what is incorrect and why.
How many ways can they sit if Jane and Mary refuse to sit next to each other and John refuses to sit on an end.
For this portion we can use the same outcomes from the first part, whatever it is and subtract the portion where John sits on the end to get the outcome of both Mary and Jane apart along with John not on the end.
How many ways can they sit of they forget their differences and sit by alternating gender (Susan, Todd, Jake).
For 3 guys and 3 girls how can I determine the outcomes? Row could look like BGBGBG or GBGBGB and then the permutations of who sits where, and again multiplied by two if they sit in the other row. This is unclear to me is it 2 * (3! + 3!) = 24 * 2 = 48 or 3! * 3! * 2 arrangements and then * 2 again for 2 rows.
We shall bring in probability to simplify the problem.
For the moment, consider Mary and Jane as blue balls, John as a yellow ball, and the other $3$ as red balls.
P(blue balls together) $=P(A)= \frac{5}{\binom62} = \frac13$
P(yellow ball) at ends $= P(B) = \frac26 = \frac13$
$P(A\cap B) = \frac13\frac{4}{\binom52} = \frac2{15}$
P(desired conditions are met) $= 1 - P(A\cup B)$
$= 1 - [P(A) +P(B) - P(A\cap B)] = 1-(\frac13+\frac13 - \frac2{15}) = \frac7{15}$
Now unrestricted permutation of three types of colored balls $= \frac{6!}{1!2!3!} = 60$
Thus valid arrangements of balls $= \frac7{15}\cdot60 = 28$
Can you now convert this into valid arrangements of people, and multiply by $2$ for the two parts of the row ?