Three straight couples a couple being respectively the lady

Three (straight) couples, a couple being respectively the lady and gentleman date, have reserved six adjacent seats at the dinner theater, determine how many ways they can sit given the ensuing.

5.1 A specific couple is already seated

5.2 Each gentleman sits adjacent to a lady other than his date

5.3 Persons of the same sex sit in the first two seats

5.4 Two ladies are identical twins, are identically addressed, and are indistinguishable

5.5 Four people sit together in a group

Solution

5.1) As a specific couple is already seated - 2 seats are taken

There are 4 seats available

No of ways in which these 4 seats can be taken - 4! = 24

5.2) Assume the following:

Case 1) When one non pair sits together first (L1 M2)

No of ways they can sit starting with Lady and Male respectively

L1 M2 L2 M3 L3 M1 M2 L1 L2 M3 L3 M1

L1 M2 L2 M3 M1 L3   M2 L1 L2 M3 M1 L3

L1 M2 M3 L2 L3 M1   M2 L1 M3 L2 L3 M1

L1 M2 M3 L2 M1 L3   M2 L1 M3 L2 M1 L3

In the case starting with the lady 2 cases are such that the dates sit next to each other and 2 ways where they dont

In the case starting with the male 1 case is such that the dates sit next to each other and 3 ways where they dont

So total cases = 5

Now we will have 5 cases if we start with couple (L1 M3)

Now we will have 5 cases if we start with couple (L2 M1)

Now we will have 5 cases if we start with couple (L2 M3)

Now we will have 5 cases if we start with couple (L3 M2)

Now we will have 5 cases if we start with couple (L3 M1)

Hence total cases = 5*6 = 30

5,3) Assuming we start with male then no of ways in which first seat can be taken = 3

No of ways in which second seat can be taken = 2

No of ways in which the rest seats can be taken = 4!

This will be same for females as well

Total ways = 3 * 2 * 4! * 2 = 288

5.4) Assume the identical person is seated first

No of ways of seating all 6 people = 6!

Exclude cases where the twins can replace each other

When one twin sits first, no of ways second can sit = 5

No of swaps possible = 5

When one twin sits second, no of ways second can sit = 5

No of swaps possible = 5 - 1 = 4 (Since one such swap is accounted for previously)

Hence so on..

Total swaps = 5 + 4 + 3 + 2 + 1 =15

Total ways = 6! - 15

5.5) consider four people as 1

No of ways of selecting 4 people = 6C4 = 5

No of ways = 5 * 3! * 4! (as 4 people can rearrange amon themselves)