In how many different ways can 4 ladies and 4 gentlemen be : GMAT Problem Solving (PS)
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# In how many different ways can 4 ladies and 4 gentlemen be

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

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04 Oct 2010, 05:40
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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
[Reveal] Spoiler: OA
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Re: Ways to sit around the table [#permalink]

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04 Oct 2010, 07:11
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eladshush wrote:
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 $$4!*4!=576$$.

Answer: C.
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Re: Ways to sit around the table [#permalink]

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04 Oct 2010, 07:32
Good question... Yet another good explanation from the Master.....
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Re: Ways to sit around the table [#permalink]

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04 Oct 2010, 07:58
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eladshush wrote:
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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Re: In how many different ways can 4 ladies and 4 gentlemen be [#permalink]

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21 Sep 2013, 11:57
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Re: In how many different ways can 4 ladies and 4 gentlemen be [#permalink]

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16 Oct 2014, 00:46
Hello from the GMAT Club BumpBot!

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Re: In how many different ways can 4 ladies and 4 gentlemen be [#permalink]

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16 Oct 2014, 01:06
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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29 Aug 2016, 01:17
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Re: In how many different ways can 4 ladies and 4 gentlemen be   [#permalink] 29 Aug 2016, 01:17
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# In how many different ways can 4 ladies and 4 gentlemen be

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