Around the World in 80 Questions (Day 2): At a dinner party, 4 married

First place can be selected in 2c1 ways and its adjacent place is selected in only one manner.
Similarly, third place is selected in 2c1 ways.

so total no.of ways people are selected= 2x2c1 = 4

They can be arranged in 4 ways.

Hence total no.of arrangements = 4x4 = 16

M1 _ M2 _ M3_ M4 _

Assume its a circular arrangement
Place women in the gaps. Each women has 2 ways.
Men can be arranged in 3! ways. Hence 6x2=12
A)

Refer to the diagram, the arrangement of no two men together can be done in this way

Let's place men first

Men can be placed in (4-1)! = 3! ways

Now when we are placing women

The first lady has the option of occupying 2 positions that are opposite to her husband

The rest of the ladies can only occupy one position each

You can try with various combinations

So total possible arrangements = 3! * 2 = 12

Ans A

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At a dinner party, 4 married couples are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group, if no two men should occupy adjacent seats and no husband and wife should occupy adjacent seats ?
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Here , we basically need to use combinatorics knowledge.

The number of arrangements of n distinct objects in a row is given by n!
The number of arrangements of n distinct objects in a circle is given by (n−1)!

If we assume that, there is no constraint about ''no two men should occupy adjacent seats and no husband and wife should occupy adjacent seats''.
R=n!/n=(n-1)!
R=(8-1)! , but there are more stuff in it. So, we basically ask for an expert reply))

At a dinner party, 4 married couples are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group, if no two men should occupy adjacent seats and no husband and wife should occupy adjacent seats ?


no 2 men and couple should sit together.

Suppose placing the men in the circle - then 3! and placing women = 3!
thus 3!*3! = 6*6 = 36.

Couple thus 3!*2 = 6*2 = 12
thus 36 - 12 = 24

ans = 24

The correct answer is 12.
Let us assume we distribute the 4 men on 4 corners. There are 3! ways of doing so.
Now we can adjust women between spaces in such a way that do not sit beside their husbands. There are two ways of doing so: W4,M1,W3,M4,W2,M3,W1,M2,W4 and M1,W2,M4,W1,M3,W4,M2,W3,M1
Hence for each arrangement of men, there are two arrangements that women can follow. Hence: 3!*2=12.

Answer: A (12)

3! ways to arrange men first. Then each position between them can be occupied by two women i.e. there are two possible women choices to sit between them. Once the woman is chosen for any seat, the rest seats will automatically be fixed.

Hence 3!*2=12

Call 4 men A,B,C,D
Name 4 women 1,2,3,4
A & 1 are couple; similarly, B2, C3, D4 are couple.

There are 3! =6 ways to arrange 4 men into 4 seats in circular table.
For each arrangement let arrange the seats for women
Take one set as example
A_B_C_D_
how many ways to fill 4 women in the blank between men such that no two men should occupy adjacent seats and no husband and wife should occupy adjacent seats. I count it manually (quite easy: between A&B only can 3 or 4; between B&C only 1 or 4; between C&D only 1 or 2; between D &A is the final one)
there are only 2 possible arrangements:
A3B4C1D2
A4B1C2D3

Thus the total arrangements = 6 * 2 = 12
A

Let married couples be M1 W1, M2 W2, M3 W3 and M4 W4 respectively.



First fix position of Men such that no two men are adjacent. Number of ways for this arrangement = (4-1)! = 6.

Now for W1 only two possible positions exist such that she does not sit adjacent to her husband M1. Similarly for W2, W3 and W4 only two possible distinct arrangement is possible as shown above.

So total number of arrangements = 6 * 2 = 12.

IMO A.

Let husbands be A,B,C & D, and their wives be 1,2,3,4 respectively.

Now let us place husbands in alternative positions, for fulfilling second requirement,

A_B_C_D_

Being circular problem the blank after D is actually between A&D

Now according to first condition following options can fill the blanks in order,

3,4

4,1

1,2

2,3

But now there is one more twist, that not all the possible combinations from this set can work for example if we choose 3 for first blank, then 1 for second and, 2 for third then we cannot fill the last position.

So, if we choose 3 for first position, then we have to choose 4 for second, 1 for third and 2 for forth.

Same holds for choosing 4 for first position.

Now we can position husbands in 3! position in a circular table.

So total number of ways are 2*3! = 12

Four men can be arranged in 3! ways.

Say we have an arrangement such as this.

------------ M1 -------------

---- M2 ---------------M3------

........- (1).....................(2).....

--------------M4----------

M1s wife can either take position 1 or position 2

Once one position is taken, other positions can be filled only in one way.

Total ways = 3! * 2 = 12

IMO A

I think B is the answer.

The Answer is E =132 ways
There will be 7! ways to arrange all the 4 couples and 132 ways taking all the constraints in the question.

At a dinner party, 4 married couples are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group, if no two men should occupy adjacent seats and no husband and wife should occupy adjacent seats ?

lets make a circular ring, hence we have

first point = 4 Option
Second Point = 3 Option
Third Point = 2 Option
Fourth Point = 1 Option
Fifth Point=1 Option
Sixth Point =1 Option

Posibble Seating arrnagement = 4*3*2*1 = 24

Option D is correct

To solve this question, we can use the concept of permutations with some restrictions. Let's break it down step by step:

Step 1: Seat the women
Since no husband and wife should occupy adjacent seats, we need to seat the women first. There are 4 women, and since the table is circular, we can arrange them in (4-1)! = 3! = 6 ways.

Step 2: Seat the men
Now, let's seat the men. Since no two men should occupy adjacent seats, we need to place them in alternate positions around the table. There are 4 men, and we can arrange them in (4-1)! = 3! = 6 ways.

Step 3: Combine the arrangements
Finally, we multiply the number of arrangements from Step 1 and Step 2 since these arrangements are independent of each other.

Total number of different possible seating arrangements = 6 (women arrangements) * 6 (men arrangements) = 36.

However, we need to consider that the table is circular, and arranging people in a circular arrangement can result in duplicate arrangements. Since there are 8 people (4 couples) and the table is circular, we have 8 equivalent starting positions for each arrangement.

So, the total number of different circular seating arrangements = 36 (total arrangements) / 8 (equivalent starting positions) = 4.

The final answer is A. 12.

-We will arrange the 4 men first=(4-1)!
-Now we must arrange the 4 women. Looking at the image, we can see that W1 can only be arranged in 2 spots, and for each spot, we can have only 1 combination that satisfies the condition=>the total number of ways to arrange 4 women=2!
=>The total arrangement=\( (4-1)!*2!\)=12
The answer is A

no two men should occupy adjacent seats - therefore, men and women are to be seated alternately. lets say we choose 4 alternate spots in 8 spots. We can place N men in distinct ways in a circular table such that they have relatively different orders in \((N-1)!\) ways. Here N=4. Therefore, that husbands can be seated in 6 ways.

in the remaining 4 alternate spots, wives are to be seated. Now lets these 4 spots are "Second 4". Husbands are in "First 4". We can switch this ie., husbands in "second 4" and wives in "first 4". so this gives 2 ways.

also, given no husband and wife should occupy adjacent seats. Once u fill 4 alternate seats with people of 1 gender, whats left is 4 other atlernate seats. So it can either be H1-W3-H2-W4-H3-W1-H4-W2-(BACK TO H1) or H1-W4-H2-W1-H3-W2-H4-W3-(BACK TO H1. 2 ways

Therefore, 6*2*2 = 24 ways.