Answer to Question #235032 in Statistics and Probability for Ndilloh

Question

Let [A1, A2], [B1, B2], [C1, C2], [D1, D2] be 4 married couples who attended a concert in Windhoek.
a) In how many ways can these 4 married couples be seated in a row of 8 seats if there are no restrictions
as to where the 8 people can sit?
b) In how many ways can these 4 married couples attending a concert be seated in a row of 8 seats if each
married couple is seated together?
c) Show that the number of ways to seat the 4 married couples attending a concert is equal to 1 152 if
and only if members of the same sex are all seated next to each other in the 8 row seats?

Answer 1

a) If there are no restrictions as to where the 8 people can sit, then the first seat can be occupied in 8 ways, the second - in 7 ways, the third - in 6 ways, etc. The total number of ways equals to $8\cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=8!=40320$.

b) Let us consider each married couple as one object. Then these objects can be placed in one row in $4!=24$ ways. But since there are 2 ways to order people in a pair (MW or WM), then the total number of ways is equal to $4!\cdot 2^4=24\cdot 16=384$.

c) If members of the same sex are all seated next to each other in the 8 row seats, then the first 4 seats are occupied by women and the last 4 seats are occupied by men, or the first 4 seats are occupied by men and the last 4 seats are occupied by women (this gives us 2 ways to choose). After that we can rearrange 4 women in their places in $4!=24$ ways, and 4 men in their seats in $4!=24$ ways. Therefore, the total number of ways equals to $2\cdot 4!\cdot 4!=2\cdot 24^2=1152$.