Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated
(a) with no restrictions?
8!=40320
(b) if each couple is to sit together?
2(4!)=48
(c) if all the men sit together to the right of all the women?
4!+4!=48
I just want to make sure that they're right. Thanks
I see why b and c are wrong now. For B, I thought it was enough to multiply by 2, but after thinking about it, it makes sense that I have to multiply by 2 four times (once of each couple).
C was more of a slight mental error on my part.
Thank you for the help