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

Question

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

such that no two men or women are together.

Bunuel · Accepted Answer

(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.

sag · Answer

No of way in which 5 men and 6 women can be seated in a row = 6! * 5! as first has to be a woman then man then woman then man n so on...

meghash3 · Answer

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 · Answer

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).

GMATGuruNY · Answer

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.

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! .

SwethaReddyL · Answer

hi  Bunuel ,
shouldn't the answer for the first question be 11! ? we are arranging 11 different people in a row

Bunuel · Answer

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.

onlyPlanA · Answer

I tried solving the first part of this question like this-

There are 11 people, so they can be arranged in 11! ways.
But, we don't want two men or women to sit together. So, taking the following pairs into consideration:

{MM}, {MM}, {M}, {WW}, {WW}, {WW}

My answer was: 11! - 6! * 2!*5

What is wrong with this approach?

Bunuel · Answer

Your subtraction term does not count all invalid arrangements. It counts only arrangements that can be divided into exactly two MM blocks, one single man, and three WW blocks.

However, invalid arrangements can have many other forms. For example,   WW  M  W  M  W  M  W  M  W  M   is invalid because two women are together, but it has no MM block and only one WW block. Also, a group such as MMM contains overlapping MM pairs, so the same arrangement may be counted more than once.

Therefore, some invalid arrangements are missed and others are counted multiple times, so this subtraction method does not work.

onlyPlanA · Answer

Ah, I get it now... Thank you!

MrMaxMan · Answer

You can solve the first problem by noticing first that there is one more woman than men. This means that for no women or men to be seated next to each other, then you must arrange them as:
W M W M W M W M W M W
We have 6 women and 5 men, using FCP we can find the number of ways by multiplying the number of options for each position.
6*5*5*4*4*3*3*2*2*1*1
Notice this is 6!*5!

For the second question, just visualize the table. If you place all the women first, there would be 6 of them with 6 spaces between them and only 5 men to fill, this means that it is impossible to have two women not sitting next to each other in a circular configuration.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards