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# In how many different ways can 4 ladies and 4 gentlemen be

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Intern
Joined: 01 Sep 2010
Posts: 19
In how many different ways can 4 ladies and 4 gentlemen be  [#permalink]

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04 Oct 2010, 06:40
6
15
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Difficulty:

25% (medium)

Question Stats:

72% (01:23) correct 28% (01:48) wrong based on 351 sessions

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In how many different ways can 4 ladies and 4 gentlemen be seated at a round table so that all ladies sit together?

A. 70
B. 288
C. 576
D. 10,080
E. 20,160
Math Expert
Joined: 02 Sep 2009
Posts: 58421
Re: Ways to sit around the table  [#permalink]

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04 Oct 2010, 08:11
3
10
In how many different ways can 4 ladies and 4 gentlemen be seated at a round table so that all ladies sit together?

A. 70
B. 288
C. 576
D. 10,080
E. 20,160

Glue the ladies together so that they create one unit, so we would have 5 units: {M1}, {M2}, {M3}, {M4}, and {W1,W2,W3,W4} --> # of different arrangements of $$n$$ objects around the table (circular arrangements) is $$(n-1)!$$, so our 5 objects can be arranged in $$(5-1)!=4!$$ ways.

On the other hand 4 women within their unit also can be arranged in 4! ways --> total $$4!*4!=576$$.

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Manager
Joined: 07 Oct 2006
Posts: 55
Location: India
Re: Ways to sit around the table  [#permalink]

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04 Oct 2010, 08:32
Good question... Yet another good explanation from the Master.....
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Joined: 02 Sep 2010
Posts: 724
Location: London
Re: Ways to sit around the table  [#permalink]

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04 Oct 2010, 08:58
1
In how many different ways can 4 ladies and 4 gentlemen be seated at a round table so that all ladies sit together?

A. 70
B. 288
C. 576
D. 10,080
E. 20,160

Treat the 4 ladies as one object, now you have 5 objects to arrange around a table (m1,m2,m3,m4,women). This can be done in (5-1)! ways
And there are 4! ways to arrange ladies among themselves

Answer = (4!)^2 = 576 or C
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Joined: 30 Mar 2013
Posts: 101
Re: In how many different ways can 4 ladies and 4 gentlemen be  [#permalink]

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16 Oct 2014, 02:06
For concept's sake, if we were to do this the opposite way, how would we do it? Say we have (8-1)! of arranging without any conditions. Then it should be 7! - number of ways 2 women can sit together - number of ways three can sit together.

so for number of ways two can sit together I get: (4-1)! and then 4C3 (in how many ways can we place 3 women in 4 slots, since I tied two together * 2)

Number of ways 3 can sit together= seat the men in (4-1)! ways. * 4C2 (in how many ways can two women be placed in 4 slots, since I tied three women together this time)*3! (for the number of arrangements of three women ties together)

This doesn't give me the correct answer. Where have I gone wrong?
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Re: In how many different ways can 4 ladies and 4 gentlemen be  [#permalink]

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05 Mar 2019, 06:52
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Re: In how many different ways can 4 ladies and 4 gentlemen be   [#permalink] 05 Mar 2019, 06:52
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