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

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

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.

Arrange the 4 men around the table. Since the table is circular, we have (4-1)! = 3! = 6 ways to do this.
Next, arrange the 4 women. Each woman has 2 choices of seats (to the left or right of her husband), but the last woman only has 1 choice (the remaining seat). So, there are 2*2*2*1 = 8 ways to arrange the women.
Multiply these together to get the total number of arrangements: 6*8 = 48.
However, this includes arrangements where two men are adjacent, which is not allowed. So we subtract the number of arrangements where two men are adjacent.
If two men are adjacent, we can treat them as a single entity. Then, we have 3 entities (the pair of men and the other 2 men) to arrange, which can be done in (3-1)! = 2! = 2 ways. The pair of men can be arranged in 2 ways (since they can switch seats), and the women can be arranged in 8 ways as before. So, there are 2*2*8 = 32 arrangements where two men are adjacent.
Subtract this from the total to get the final answer: 48 - 32 = 16.
So, the answer is C

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)

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.

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

-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

We need to ensure two conditions here:-
(1) No two men should occupy two adjacent seats
and (2) No husband and wife should occupy adjacent seats.

In the table with 8 seats, we will first seat the men on every alternate seat as shown in the fig. 1, and later fill women in empty seats between men. Now lets consider M1, M2, M3, and M4 be the four men and W1, W2, W3, and W4 be the four women. We are assuming that M1 and W1 are couples and similarly for others.

Initially for seating four men on a circular table:- First man will have only 1 choice, as all seats are empty and his choice wont create any new pattern, second man will have remaining 3 choices, then third man will have 2 and last man will have only 1 choice. Or we can also use the formula for circular arrangement as:- (n-1)! = (4-1)! = 6 ways.

Now for seating women we need to ensure no husband and wife are seating adjacent. For this we will assume one of the six arrangements of men as shown in the fig. 2. Now for seating W1 we have only two options, W1 can seat either between M3 and M4 or between M3 and M2. I have drawn both the cases in fig. 2. Once we fix the position of W1 in one of the above two places, all other women's position are automatically fixed.
So we can conclude that, for any given arrangement of men, we can sit the women in two ways.

Hence the total ways of arrangement are:- 6*2 = 12 ways.

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 ?

A. 12
B. 14
C. 16
D. 24
E. 132

Solution:
If M1 is seated then the adjacent chairs can be filled in 3C2 ways(two out of W2,W3,W4 can sit), after that the adjacent chairs can be filled in 2C1 (1 out of M3 and M4 can sit) ways and then again in 2C1 ways (1 out of W1 and W4 can sit).
Total no. of ways= 3C2*2C1*2C1=12 ways. ANSWER A

Consider a circular table consisting of 8 seats .

Let the four couples be M1 W1, M2 W2, M3 W3, M4 W4

We can first fill the seats by four men, who will be seating in alternate seats.

As per circular permutation , four men can be arranged in (4-1)! = 6 ways

For the sake of simplicity , lets assume the below arrangement :

... __M1__
M4..........M2
... __M3__

Now we have four women W1, W2, W3, W4.

W1 will have two choices to choose where to sit. She can choose seat between M2 and M3 or between M3 and M4

Lets say she chooses to sit between M2 and M3

... __M1__
M4..........M2
... __M3 W1

Now if we notice carefully, we can see that since both space around M4 is empty W4 can't take any of that space .
Thus W4 will have only one option to sit.

... __M1 W4
M4..........M2
... __M3 W1

Now , we have two seats available for W2 and W3. W3 can't sit beside M3 and so only one option is available for W3.

...W3 M1 W4
M4...........M2
... __M3 W1

Now W2 will take the last remaining seat.

Thus, only the woman that will sit first will have 2 choices to select the seat, rest all will have only one choice.
Since all couples are different this will always be the case in any arrangement .

So no of ways in which all women can be arranged = 2

Total arrangements= 6 * 2 = 12

There are 8 seats to be filled.
Suppose we start with men (men or women is the same).
First seat: 4 choices
Second seat (has to be a wife but not the one of the first one): 3 choices
Third seat: 2 choices (3-1 of the second seat)
Fourth seat: 2 choices
Fifth, sixth, seventh, eighth seat: 1 choice

Choices possible = 4*3*2*2
But the table is circular, in particular places that matter for the choice are four (the other four are a consequence)
Therefore: 4*3*2*2/4 = 12 -> IMO A

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 ?
4 married couple: A-A'; B-B'; C-C';D-D'.
Put A, B, C, D in 4 inadjacented seats. => Fixing the position of A, changing the positions of remains, we have arrangements.
We have 4 blanks seats for A', B', C', D' (inadjacented)
In case A-B-C-D, the seat between A&B, we can choose C' or D'.
We have 3 arrangements: C'-A'-B'-D' (rejected); C'-D'-A'-B'; D'-A'-B'-C'.
So we have 2*3!=12=>A

Let the 4 husbands be H1,H2,H3,H4.
Let the 4 wifes be W1,W2,W3,W4.

Assume the first position is filled by H1.
The next adjacent position can be filled by W2 or W3 or W4. Hence 3 possibilities.
Assume W2 has filled the 2nd position, the next position in order can be filled by H3 or H4. 2 Possibilities.
Assume H3 has been filled in the 3rd position, the 4th position can be filled by W4 or W1. 2 possibilities.
After this all the rest of the positions will be filled without any more possibilites.
Therefore, total cases = 3*2*2 = 12

Option A.

Given: 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 ?

Let Women be denoted by W1, W2, W3, W4 and Men by M1, M2, M3, M4
with couples being (W1,M1), (W2,M2), (W3,M3) and (W4,M4)

There are 2 restrictions:
1. no two men should occupy adjacent seats
2. no husband and wife should occupy adjacent seats

From 1st case, men can only occupy alternate seating positions i.e. 1 women has to be seated between them.
Let's seat women first.

No. of ways to arrange seating of 4 women around a circular table = (n - 1)!
= (4 -1)!
= 3!
= 6 ways

Now there are 4 positions left on the table. Now from Case 2, the husband cannot sit adjacent to his wife i.e. to the either side of his wife. That leaves 2 out of 4 positions for a husband.
This makes 2 such arrangements for each of the 6 ways of arranging the women.

Hence Total number of ways 4 couples can be seated is 6 * 2 = 12 ways

Hence Option A is the CORRECT choice