Six friends sit in adjoining seats in a row. A wall, 6 seats, an aisle, 6 more seats, then a wall "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?"
I've figured out how many ways all 6 can sit together. 6! * 2(sides of the theater)=1440
I the figured out how many ways the two friends could sit together and then subtract that amount. 5!*2!*2(sides of the theater) and end up with 1440-480=960 ways to sit.  
However, the last part of the question is confusing me.  "John refuses to sit on an end". 
Please help. 
 A: I agree with your interpretation. Seats that have an aisle between them are not adjoining. To solve: Use Inclusion/Exclusion.
Total number of orders, as you identified, is $12\times 5!$ (there are $12$ seats for the first person, but once they are selected, everyone else has to sit on the same side of the aisle, so there are $5!$ ways to place everyone else)
Number of ways for John to be at an end: $4\times 5!$ (there are $4$ end seats, but once John is placed, everyone else has to sit on the same side of the aisle, and there are $5!$ ways to seat them).
Number of ways for Jane and Mary to be next to each other: $10\times 2!\times 4!$ (There are $10$ pairs of adjacent seats that are not separated by an aisle. Once you choose a pair, choose an order for Jane and Mary, then place everyone else on the same side of the aisle in $4!$ ways)
Number of ways for Jane and Mary to be next to each other while John is at an end: $4\times 4\times 2!\times 3!$ (There are $4$ end seats to place John. Regardless of which spot you put John, there are $4$ adjacent pairs of sets on the same side of the aisle for Mary and Jane. Choose an order for them, and then there are $3!$ ways to seat the remaining three people).
Total ways with Jane and Mary separated and John not at the end:
$$12\times 5! - 4\times 5! - 10\times 2!\times 4! + 4\times 4\times 2!\times 3! = 672$$
