6 men and 6 women are seated around a table. What is the probability t

Question

6 men and 6 women are seated around a table. What is the probability that the men and women sit alternately at the table?

A. 1/462
B. 1/55
C. 3/55
D. 2/11 
E. 36/55

GMATinsight · Accepted Answer

Alernate sitting methods = 6!*5!

toral arrangements = 11!

Probability  = \frac{6!*5!}{11!}= \frac{6!*120}{11*10*9*8*7*6!} = \frac{1}{462}
 
Answer Option A

prince11goyal · Answer

Total number of  unique  arrangement possible around a circular table =  (12-1)! = 11!

Assume we have a 6 seat table to arrange 6 men. Total number of cases =  (6-1)! = 5! 
Now we can put 6 chair between those 6 men in a circle. Total number of cases to arrange women on those extra seats =  6!  (it's a circle, but now we already have a starting reference once we seat all the men, thus not 5!)

Probability = (5! * 6!)/11! = 1/462.

rvgmat12 · Answer

Total combinations = 11!

Arrange women first = 5! Then arrange men in 6! ways

Probability = 5!*6!/(11!) = 1/462

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4ce5 · Answer

Can you please clarify what you meant about having a starting reference? I'm not really understanding why we do 6! for the women instead of 5!

Bunuel · Answer

The arrangement of n distinct objects in a row is denoted by n!.  
  The arrangement of n distinct objects in a circle is represented by (n-1)!.

The distinction between arranging in a row and in a circle is as follows: if we shift all objects by one position, we achieve a different arrangement in a row, but maintain the same relative arrangement in a circle. Therefore, for the number of circular arrangements of n objects, we have:

\frac{n!}{n} = (n-1)! .

To demonstrate this, let's consider a simpler case: {1, 2, 3} arrangement in a row is distinct from {3, 1, 2} and from {2, 3, 1} arrangement.

However, the below three arrangements in a circle are identical:

https://gmatclub.com/forum/download/file.php?id=71640

Next, if we place three letters a, b, and c between these numbers, shifting them by one position would no longer yield the same arrangement. This is because it now matters whether a is positioned between 1 and 3 or between 1 and 2:

https://gmatclub.com/forum/download/file.php?id=71641

Therefore, the number of arrangements of a, b, and c between 1, 2, and 3 is now 3!, rather than 2!, due to the relative positioning of these letters in regards to the numbers.

6 men and 6 women are seated around a table. What is the probability that the men and women sit alternately at the table?

A. 1/462
B. 1/55
C. 3/55
D. 2/11
E. 36/55

According to the above, the number of ways to seat 12 people, consisting of 6 men and 6 women, around a table without any restrictions is (12 - 1)! = 11!.

To arrange them alternately, we calculate (6 - 1)! = 5! for one group, and arrange the members of the other group between them in 6! ways.

Therefore, the probability is 5!6!/11!= 5!/(7*8*9*10*11) = 1/(7*3*2*11) = 1/462.

Answer: A.

Hope it's clear.

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6 men and 6 women are seated around a table. What is the probability t

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6 men and 6 women are seated around a table. What is the probability t

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