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

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Re: Around the World in 80 Questions (Day 2): At a dinner party, 4 married [#permalink]
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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
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Re: Around the World in 80 Questions (Day 2): At a dinner party, 4 married [#permalink]
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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.
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Re: Around the World in 80 Questions (Day 2): At a dinner party, 4 married [#permalink]
Bunuel wrote:
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

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over \$20,000 in prizes: Courses, Tests & more

lets have 8 places _ _ _ _ _ _ _ _
Lets start with the men. The they can sit in 4! ways: slots 1st, 3rd, 5th, 7th. but because, they sit around we can move each husband a position to the right, and they have the same relative position. So we divide by 4. The number of positions for the men is 4! / 4 = 6.
The for each position of a man, his wife can have only two possible places, next to the man opposite to husband. So we multiply by 2.
6x2 = 12
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Re: Around the World in 80 Questions (Day 2): At a dinner party, 4 married [#permalink]
Bunuel wrote:
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

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over \$20,000 in prizes: Courses, Tests & more

We first place the men in alternate positions in the sequence - M1, M2, M3 and M4 - this sequence can be decided in (4 - 1)! = 6 ways
(the couples are M1 W1 M2 W2 M3 W3 M4 W4)
The possible positions for the women are also shown.

Attachment:

11.png [ 5.9 KiB | Viewed 496 times ]

If the top left position (north-west) is W2, the bottom left (south west) is W1, bottom right (south east) is W4 and top right (north east) is W3
Similarly: If the top left (north-west) is W3, the top right (north east) is W4, the bottom right (south east) is W1 and bottom left (south west) is W2
Thus, there are 2 ways in which the women can be placed without being seated beside their husbands.

Thus, total ways = 6 * 2 = 12