In how many ways can 4 married couples [total of 8 people] be seated in a row if: [a] there are no restrictions on the seating arrangement? [b] persons A and B must sit next to each other?[c] there are 4 men and 4 women and no 2 men or 2 women can sit next to each other? [d] there are 5 men and they must sit next to one another? [e] there are 4 married couples and each couple must sit together?
Solution:
Given, there are 4 married couples
Total number of people = 8
we have to find how many ways can 4 married couples be seated in a row
a] if there are no restrictions on seating arrangement
The number of possible ways = n!
Here, n! = 8!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320 ways.
Therefore, there are 40320 ways that the people can be seated when there is no restriction on the seating arrangement.
b] if persons A and B must sit next to each other
The possible ways = 7! × 2!
= [7 × 6 × 5 × 4 × 3 × 2 × 1] × [2 × 1]
= 5040 × 2
= 10080
Therefore, if persons A and B must sit next to each other there are 10080 ways of seating arrangement.
c] if there are 4 men and 4 women and no 2 men or 2 women can sit next to each other
This implies the restriction of having persons of opposite sex next to each other.
We can have any one of the 8 persons in the first position.
So, both man and woman can be arranged without having same next to
each other = 8 × 4 × 3 × 3 × 2 × 2 × 1 × 1
= 1152 ways
Therefore, if there are 4 men and 4 women and no 2 men or 2 women can sit next to each other, there are 1152 ways of seating arrangement.
d] if there are 5 men and they must sit next to one another
There are 5 men and 3 women. 5 men sit next to one another.
The possible number of ways = 5! × 4!
= [5 × 4 × 3 × 2 × 1] × [4 × 3 × 2 × 1]
= 120 × 24
= 2880 ways
Therefore, if there are 5 men and they must sit next to one another, there are 2880 ways of seating arrangement.
e] if there are 4 married couples and each couple must sit together
4 pairs of couples can be arranged in 4!
Each couple can be arranged in 2! Ways.
The possible number of ways = 2! × 2! × 2! × 2!
= [2× 1] × [2 × 1] × [2 × 1] × [2 × 1] × [4 × 3 × 2 × 1]
= 2 × 2 × 2 × 2 × 24
= 16 × 24
= 384 ways
Therefore, if there are 4 married couples sitting together, there are 384 ways of seating arrangement.
In how many ways can 4 married couples [total of 8 people] be seated in a row if: [a] there are no restrictions on the seating arrangement? [b] persons A and B must sit next to each other?[c] there are 4 men and 4 women and no 2 men or 2 women can sit next to each other? [d] there are 5 men and they must sit next to one another? [e] there are 4 married couples and each couple must sit together?
Summary:
The possible number of ways can 4 married couples [total of 8 people] be seated in a row if:
[a] there are no restrictions on the seating arrangement is 40320 ways
[b] persons A and B must sit next to each other is 10080 ways
[c] There are 4 men and 4 women and no 2 men or 2 women can sit next to each other in 1152 ways.
[d] there are 5 men and they must sit next to one another is 2880 ways
[e] there are 4 married couples and each couple must sit together in 384 ways.
There are $6!$ possible seating arrangements. From these, we must exclude those in which one or more couples sit in adjacent seats.
There are three ways to select a couple who sit in adjacent seats. That gives us five objects to arrange, the couple and the other four people. The objects can be arranged in $5!$ ways. The couple that sits together can be arranged internally in $2!$ ways. Hence, there are $$\binom{3}{1}5!2!$$ seating arrangements in which a couple sits in adjacent seats.
However, if we subtract these seating arrangements from the total, we will have subtracted too much since we have counted seating arrangements in which two couples sit together twice, once for each way we could designate one of the couples as the couple that sits in adjacent seats. Since we only want to subtract such couples once, we must add them back.
There are $\binom{3}{2}$ ways to select two couples that sit together. That gives us four objects to arrange, the two couples and the two other people. The objects can be arranged in $4!$ ways. Each of the two couples that sit in adjacent seats can be arranged internally in $2!$ ways. Hence, there are $$\binom{3}{2}4!2!2!$$ seating arrangements in which two couples sit together.
When we subtracted arrangements in which a couple sits together, we counted seating arrangements in which all three couples sit together three times, once for each way we could have designated one of those couples as the couple that sits together. When we added arrangements in which two couples sit together, we counted seating arrangements in which all three couples sit together three times, once for each of the $\binom{3}{2}$ ways we could have designated two of the three couples as the ones that sit together. Therefore, we have not excluded seating arrangements in which all three couples sit together at all.
There are $3!$ ways to arrange three couples. Each couple can be arranged internally in $2!$ ways. Hence, the number of seating arrangements in which all three couples sit together is $$\binom{3}{3}3!2!2!2!$$
By the Inclusion-Exclusion Principle, the number of seating arrangements of the three couples in which no couples sit together is $$6! - \binom{3}{1}5!2! + \binom{3}{2}4!2!2! - \binom{3}{3}3!2!2!2!$$
The probability that no couple sits together is $$\frac{6! - \dbinom{3}{1}5!2! + \dbinom{3}{2}4!2!2! - \dbinom{3}{3}3!2!2!2!}{6!} = 1 - \frac{\dbinom{3}{1}5!2! - \dbinom{3}{2}4!2!2! + \dbinom{3}{3}3!2!2!2!}{6!}$$