Question

Two pairs of siblings (two brothers and a brother/sister), and three only children (two men and one woman) sit in a row of seven consecutive seats. How many ways can they be seated:
i. with no restrictions;
ii. alternating genders;
iii. such that the women are all consecutive;
iv. such that twin siblings sit next to one another?

Answer 1

Total number of people = 2 + 2 + 3 = 7

(i) If there is no restriction then the number of ways of seating 7 people in 7 seats = 7! = 5040 ways

(This is by the rule which states that the number of arrangements of n distinct thing all taken together, without repetition = n!.)

We can also look at the box method below. The 1st seat can be taken by any of the 7 people, the second place by abny 6, the 3rd place by any 5 and so on...the last seat can be filled by the last remaining person.

Seat	1	2	3	4	5	6	7
# of Persons	Any 7	Any 6	Any 5	Any 4	Any 3	Any 2	Any 1

Therefore Total ways = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.

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(ii) Alternating genders. There are 5 men and 2 women. There is 0 possible combinations where men and women can sit alternatively. eg M W M W M M M. We see that the last 3 are all men, which does not adhere to the requirement of alternationg. Therefore 0.

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(iii) Basically we are saying that the women should be together. Consider them as 1. So there are 5 men and the group of 2 women considered as 1. Therefore the number of ways of arranging

1	2	3	4	5	6
M	M	M	M	M	W1W2

these 6 are 6! ways, but we must remember that the 2 women can themselves be placed in 2! ways(As W1W2 and W2W1). Therefore total number of ways = 6! * 2! = 720 * 2 = 1440

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(iv) Lets now group the twin siblings. We have the 2 brothers as the 1st set, the brother/sister as the second set and then 3 individual people, which make it 5. Therefore the number of arrangements = 5!. But as done in (iii) the sibling in each set can sit in 2! ways, therefore the total number of arrangements = 5! * 2! * 2! = 120 * 2 * 2 = 480

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