6 friends go to the movies and sit in adjoining seats in one row. The theater is configured so that in any row there is a wall, 6 seats, an aisle, 6 more seats, and then a wall.

As we know that Jane, Mary, John Susan, Todd and Jake. 6 friends.

a) How many ways can they sit if Jane and Mary refuse to sit next to each other?

b) 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.

c) How many ways can they sit of they forget their differences and sit by alternating gender (Susan, Todd, Jake).

Solution Jane , mary , John , Susan , Tod . Jake - wall aisle wall . Total number of ways to fill sits in adjoining seat = 6 1 +61 = 2x61 = 2 x ( 6 x 5 x 4 x3x2X1 ) = 1440 a) . Jane and Mary refuse to sit next to each other - 2 X 6 4 4 3 2 1 = 2 x 6X4 x4X3X2X| = 1152 ways ( b). Jane and many refuse to sit next to each other and John refuse to sit on an end. pattern = 2x 5 4 4 3 2 1 ways = 2 x 5 x 4x 4 x 3X 2X 1 = 960( c ) . sitting by alternate gender. . . make = 3 , female= 3 patterns : () 2 X MAMFMF 2 x (3 x3 X 2X2X IXI) 72 . 2 X F J M F M F m 2x (3 X 3 x 2 X 2 X 1X 1 ) . 7 2 . .. total ways = 72+72 ' = 144