|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 16 Apr 2010
Posts: 225
Followers: 3
Kudos [?]:
29
[0], given: 12
|
Seven men and five women have to sit around a circular table [#permalink]
 28 Jul 2010, 19:53
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
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?
|
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[1] , given: 829
|
Re: Arrangement In a Circle [#permalink]
29 Jul 2010, 02:41
1
This post received
KUDOS
|
|
|
|
|
|
Intern
Joined: 03 Mar 2010
Posts: 41
Followers: 1
Kudos [?]:
6
[0], given: 1
|
Re: Arrangement In a Circle [#permalink]
29 Jul 2010, 01:57
It is a tough question, at least for me  So let me try, sure that the following is not the most simple way. I first arrange 5 women around the table, there are 5!/5=4! ways. Then I select randomly 5 men and insert them, again randomly, into places between women, there are C_5_7*5! ways. OK, now is the last step, it remains 2 men, and I can insert them to any places as I like. I distinguish 2 cases: - Case 1: 1 set of 3 men together, then I have to insert the 2 remaining men into the same segment limited by 2 consecutive women. There are 10 places for the first man then 3 for the second, meaning a total of 10*3 ways. Nevertheless, it is remarked that considering the equivalent roles of these three men, I will divide the result of this case by 3. - Case 2: 2 sets of 2 men together, then I insert the 2 remaining men into 2 different segments limited by 2 consecutive women. There are 10 places for the first man and then 8 for the second, meaning a total of 10*8 ways. Again, it is remarked that the result of this case should be divided by 4. In conclusion, the total number of ways is: 4!*C_5_7*5!*[10*3/3+10*8/4]=1814400 ways.
_________________
Hardworkingly, you like my post, so kudos me.
|
|
|
|
|
|
Manager
Joined: 16 Apr 2010
Posts: 225
Followers: 3
Kudos [?]:
29
[0], given: 12
|
Re: Arrangement In a Circle [#permalink]
29 Jul 2010, 02:42
Thanks Bunuel and will definitely do in future posts
|
|
|
|
|
|
Intern
Joined: 15 May 2010
Posts: 3
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Arrangement In a Circle [#permalink]
31 Jul 2010, 08:11
Dear Bunuel,
Can you explain why in a circle it is (n-1)!
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[0], given: 829
|
Re: Arrangement In a Circle [#permalink]
31 Jul 2010, 19:40
|
|
|
|
|
|
Manager
Joined: 02 Apr 2010
Posts: 102
Followers: 1
Kudos [?]:
2
[0], given: 1
|
Re: Arrangement In a Circle [#permalink]
01 Aug 2010, 09:22
Think the mathematic working is reasonable. But how do we rationalize it from normal logic? My thinking was:- if no women(W) are to be together, they will have to be partitioned by men(M). So I paired them up in this manner:-
MW MW MW MW MW MM
So the number of ways to arrange the pairs is 6 !
Next to address the pairing:-
In first pair : 7C1 x 5C1 = 7 x 5 = 35
2nd pair : 6C1 x 4C1 = 6 x 4 = 24
3rd : 5 x 3 = 15
4th : 4 x 2 = 8
5th : 3 x 1 = 3
6th :2C1 x 2C1 = 4
Hence total is 6! x (35 + 24 + 15 + 8 +3 + 4) = 64, 080.
Will anyone be able to point out to me where I've gone wrong?
Thks.
|
|
|
|
|
|
Senior Manager
Joined: 16 Apr 2009
Posts: 349
Followers: 1
Kudos [?]:
25
[0], given: 10
|
Re: Arrangement In a Circle [#permalink]
07 Aug 2010, 15:23
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 ?
_________________
Always tag your question
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[0], given: 829
|
Re: Arrangement In a Circle [#permalink]
07 Aug 2010, 15:26
|
|
|
|
|
|
Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 694
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Followers: 13
Kudos [?]:
50
[0], given: 15
|
Re: Arrangement In a Circle [#permalink]
07 Aug 2010, 16:56
Bunuel wrote: jakolik wrote: 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? # 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. 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
_________________
Consider kudos, they are good for health
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[0], given: 829
|
Re: Arrangement In a Circle [#permalink]
08 Aug 2010, 01:55
mainhoon wrote: Bunuel wrote: jakolik wrote: 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? # 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. 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 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.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 694
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Followers: 13
Kudos [?]:
50
[0], given: 15
|
Re: Arrangement In a Circle [#permalink]
08 Aug 2010, 08:07
I understand why we use the (n-1)! for the circular case. What I was confused on is this: I guess then that the men get to use the (n-1)! = nPn/n because they go in FIRST. In the case of the women we use the nPr = nCr r! because they go in NEXT, so in this case the order in which group goes in matters. In the original question with 7Men and 5Women, the answer is 6! 7C5 5!, with 7Men and 7Women, the answer would be 6! 7C7 7! <- because the men that are already seated now influence the arrangement of the women (can't use 6! for women). And that would be the same answer whatever order they go in. - Now back to the case of 7M and 5W, we solved it first seating the Men to get the 6!. Can we solve it FIRST seating the women and then men, it should be the same answer? - What would be the general answer in this case: C chairs M men and W women (where C >= M+W) and no 2 men or women sit next to each other?
_________________
Consider kudos, they are good for health
|
|
|
|
|
|
|
Re: Arrangement In a Circle
[#permalink]
08 Aug 2010, 08:07
|
|
|
|
|
|
|
|
|
|
|
close
