Math
posted by Monique .
Four couples have purchased ticket to the school play. Their sear are next to one another in a single row. If each couple sits side by side, how many seating arrangements are possible?
(I came up with 24. correct??)

I count 64 posibilitys: p=person s=seat
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss
p ssssssss 
is that with the requirement that each couple sit side by side?

no

if the couples have to sit side by side then i count 32

In the first seat: There are 8 possibilites for who sits there.
In the second seat: There is only 1 possibility (the other member of the couple).
In the third seat: There are 6 possibilites for who sits there.
In the fourth seat: There is only 1 possibility (the other member of the couple).
In the fifth seat: There are 4 possibilities for who sits there.
In the sixth seat: There is only 1 possibility (the other member of the couple).
In the seventh seat: There are 2 possibilities for who sits there.
In the eighth seat: There is only 1 possibility (the last member of the last couple).
Answer: 8*1*6*1*4*1*2*1 = 384 different possibilities! 
There are 384 possibilities.

uhhhhhhhh are you sure cause that doesn't sound posible. especialy if the couples have to sit side by side

i know u included the couples sitting side by side but still

Yes, if the couples were not required to sit next to each other then the number of possibilities would be (8*7*6*5*4*3*2*1) = 40,320. I've taken Statistics at Georgia Tech and this was the fun part!!

got it. Thanks so much.

there are six ways each of them can sit next to each of them sit. five to sit next to them. six times six = thirtysix
five times five = 30 add them together you get 66