Seven men and five women have to sit around a circular table

Question

Seven men and five women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done?

A. 86,400
B. 172,800
C. 518,400
D. 1,814,400
E. 2,721,600

Bunuel · Accepted Answer

# of arrangements of 7 men around a table is  (7-1)!=6! ;
There will be 7 possible places for women between them, 7 empty slots. # of ways to choose in which 5 slots women will be placed is  C^5_7=21 ;
# of arrangements of 5 women in these slots is  5! ;

So total:  6!*21*5!=1,814,400 .

Answer: 1,814,400.

jakolik please post PS questions in PS forum and also try to provide answer choices.

abhijeetkarkare · Answer

Dear Bunuel,
Can you explain why in a circle it is (n-1)!

Bunuel · Answer

From Gmat Club Math Book (combinatorics chapter):

"The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:

R = \frac{n!}{n} = (n-1)! "

Check Combinatorics chapter of Math Book (link in my signature).

Hope it's clear.

ichha148 · Answer

Bunuel , Can you please explain this part

There will be 7 possible places for women between them

M1_ M2_M3_M4_M5_M6_M7 - i counted 6 possible places , where did i miss ?

Bunuel · Answer

Men are placed around a circular table so there will be one more place between M7 and M1.

mainhoon · Answer

Bunuel

I understand the 6! part for the #ofmen arrangements. However for the #of women arrangements you use 7C5. I see there are 7 slots available and 5 women. 
- For the women we are essentially using nPr = nCr r! arrangements. 
- Let me extend the argument - if there were 7 women, then the #of arrangements for women by the nCr r! logic would be 7C7 7!=7!, however it would seem that we should be using the (r-1)!= 6! (similar to men) for the women too. What am I missing. Thanks

Bunuel · Answer

So basically you are saying that women are also placed in a circle and we should use  (n-1)!  formula. The reason we use  (n-1)!  for circular arrangements is that when we shift all  n  objects by one position, we still would have the same arrangement of all objects relative to each other.

But consider different situation: a table with 6 chairs where every second chair is already taken by men. Now, if we place women in empty places we would have one arrangement BUT if we shift these women by one position we would have different arrangement as relative position of 6 changed. So # of different arrangements in this case would be 3! not 2!.

Hope it's clear.

NGGMAT · Answer

Dear Bunnel

the qs only says that the Women should not sit next 2 each other... but men can right? so y are we assuming

M_M_M_M_M_M_M_ : 14 Places

we can also have:

M_M_M_M_M_MMM: 12 places
MMM_M_M_M_M_M: 12 places

in this case the answer will be:

6!*5! right??

Bunuel · Answer

Not sure I can follow you...

Those 12 people are around a circular table. So, when no 2 women are together, at least 2 men must be together.

NGGMAT · Answer

yes, so in this case the solution will be 6! * 5! right?? how is this different from your solution.... our purpose of 2 women not being together can be solved like this also??
basically where did you get 21 from in 6! * 5! * 21 solution provided by you for the above problem

Bunuel · Answer

No. Imagine 7 men around a table so that there is an empty slot between any two: Table.png We'll have 7 slots (blue dots) but we have only 5 women. Now, those 5 can take any of the 7 available slots (and in this case no two women will be together). The number of way to choose 5 out of 7 is C^5_7=21 .

JusTLucK04 · Answer

Awesome approach Bunuel...
Why is this wrong neways: 12 Guys..No of ways 11!
Take 2 Women and bind them together...Now 11 guys and no of arrangements is 10!*2!
This covers all arrangements with at least 2 women together....

No of arrangements for no women together is: 11!- 10!*2! ---> wrong answer
Why?

mittalg · Answer

The error is in 10!*2!. This is not taking into account which 2 women are together. This is always done when you know that which two persons are together. For example, if a question says that A & B should always be together, then only you can do it. Now they are not specifying any particular group of women. Hence, this approach will lead you to a wrong answer. Had there been only 2 women in the above question, then your approach would have worked well. If we had 7 men and two women and the rest of the question is same, we would get an answer as:

Approach 1: 6! * 7C2 * 2! = 7! *6

Approach 2: 8! - 7!2! = 6*7!

I hope you get my point.

If you try to apply this approach, you have to first fix those two women who are together, which can be done in 5C2 ways. But when you are arranging them, there would be a case when three women W1W2W3 are together which is included multiple number of times. You have to subtract these cases, which is a very tedious process and hence unproductive.

PiyushK · Answer

I visualize these arrangements as following: Roundarrangement.jpg

KarishmaB · Answer

Yes, men can sit together but women cannot.

Don't assume places to be empty chairs. Think of a big round table. Each person who comes and sits around the table, brings his/her own chair along. Say the 7 men come and sit around the round table. They will be able to do that in 6! ways. Now, there is space between each pair of men. How many distinct spaces are there? 7 because there are 7 men say M1, M2, M3 till M7. So now you have empty space to the right of M1 and right of M2 and right of M3 etc. The women can take any 5 of these 7 spaces. Note that 2 women cannot take the same space because two women cannot sit together.

Say, the 5 women took 5 spots each to the right of M1, M2, M3, M4 and M5. So now spaces to the right of M6 and M7 are vacant. This means M6, M7 and M1 are sitting together with no one in between them. This takes care of the cases you have pointed out. So when we select 5 of the 7 spaces, we take care of all cases.

In how many ways can 7 men sit around a circular table? In 6! ways.
In how many ways that women select 5 of the 7 distinct places and arrange themselves in those places? In 7C5 * 5! ways.

Total arrangements = 6!* 7C5 * 5!

VIVA1060 · Answer

Hi  VeritasKarishma   Bunuel

- I started with women first: Arranged in 4! ways
- Then I arranged 5 men: 7C5*5!
- Then for the remaining 2 men, we can arrange them in 10C2 men

= 4! * 7C5 * 5! * 10C2 = 2,721,600

Can you please help me understand where I am going wrong?

KarishmaB · Answer

You are double counting cases.

Say 7C5 you pick M1, M2, M3, M4 and M5.
Now, 10C2, you pick the spot to the right of M1 for M6.

Or
Say 7C5 you pick M6, M2, M3, M4 and M5.
Now, 10C2, you pick the spot to the left of M6 for M1.

You are counting these as two different cases though they are not.

Whenever you choose some from a group and then choose again from the leftover group to arrange all together, keep the double counting issue in mind.

AbhaGanu · Answer

@Bunnel how is this calculated C^5...down 7 = 21

Bunuel · Answer

7C5 = \frac{7!}{5!(7-5)!}=\frac{7!}{5!*2!}=21

21. Combinatorics/Counting Methods 
     Theory   
 Math: Combinatorics  
 Combinatorics Made Easy! 
 1 hour Combinatorics Video Lesson 
 Combinations while dealing with similar and dissimilar things 
 When Permutations & Combinations and Data Sufficiency Come Together 
     Questions   
 Advanced Topic: Probability and Combinations Questions With Solutions 
 Letter Arrangements in Words 
 Seating Arrangements in a Row and around a Table 
 Constructing Numbers, Codes and Passwords 
 DS Questions 
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For more:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

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