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meghash3
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Thanks all and Bunuel. That is the OA.

I have a doubt still..
Wont we multiply 6!5! by 2!, as there are two possibilities, one we fix women in first position and other that we fix a man in hte first position?

Bunuel
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No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is \(6!\). If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is \(5!\). So total \(6!*5!\).

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in \((6-1)!=5!\) # of ways (# of circular permutations of \(n\) different objects is \((n-1)!\)). 6 women between them can be seated in \(6!\) # of ways. Total: \(5!6!\).

Hope it helps.
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Thanks all and Bunuel. That is the OA.

I have a doubt still..
Wont we multiply 6!5! by 2!, as there are two possibilities, one we fix women in first position and other that we fix a man in hte first position?

It's only possible woman to be first and than man: W-M-W-M-W-M-W-M-W-M-W (remember there are 6 women and 5 men).
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To count circular arrangements without applying a special formula:
1. Place someone in the circle.
2. Count the number of ways to arrange the REMAINING people.

Quote:
In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?
Here, men and women must ALTERNATE.
After one of the 6 men has been placed at the table, count the number of options for each empty seat, moving clockwise around the table:
Number of options for the first empty seat = 6. (Any of the 6 women.) )
Number of options for the next empty seat = 5. (Any of the 5 remaining men.)
Number of options for the next empty seat = 5. (Any of the 5 remaining women.)
Number of options for the next empty seat = 4. (Any of the 4 remaining men.)
Number of options for the next empty seat = 4. (Any of the 4 remaining women.)
Number of options for the next empty seat = 3. (Any of the 3 remaining men.)
Number of options for the next empty seat = 3. (Any of the 3 remaining women.)
Number of options for the next empty seat = 2. (Either of the 2 remaining men.)
Number of options for the next empty seat = 2. (Either of the 2 remaining women.)
Number of options for the next empty seat = 1. (Only 1 man left.)
Number of options for the last empty seat = 1. (Only 1 woman left.)
To combine these options, we multiply:
6*5*5*4*4*3*3*2*2*1*1 = 6!5!.
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hi Bunuel,
shouldn't the answer for the first question be 11! ? we are arranging 11 different people in a row
meghash3
The number of ways in which 5 men and 6 women can be seated

1. In a row
2. Around a table, such that no two men or women are together.
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SwethaReddyL
hi Bunuel,
shouldn't the answer for the first question be 11! ? we are arranging 11 different people in a row


The number of ways in which 5 men and 6 women can be seated

1. In a row;
2. Around a table;

such that no two men or women are together.

The condition “such that no two men or women are together” applies to both cases. It just was not formatted properly in the original post. Edited now.
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