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# There are 6 gentlemen and 5 ladies to dine at a round table.

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Intern
Joined: 25 Jan 2013
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There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]

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27 Jan 2013, 21:17
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Difficulty:

35% (medium)

Question Stats:

71% (01:33) correct 29% (02:14) wrong based on 229 sessions

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There are 6 gentlemen and 5 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are seated together.

A. 85400
B. 36400
C. 86400
D. 65600
E. 88600
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Jan 2013, 23:29, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.

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Math Expert
Joined: 02 Sep 2009
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Re: There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]

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27 Jan 2013, 23:45
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antoxavier wrote:
There are 6 gentlemen and 5 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are seated together.

A. 85400
B. 36400
C. 86400
D. 65600
E. 88600

6 men can be arranged around a table in (6-1)!=5! ways.

Now, if we place each woman between the men then no two women will be together. There are 6 available slots between the men, thus 6C5=6 is the number of ways we can choose 5 slots out of 6 for the women.

Next, these 5 women can be arranged in 5! ways, hence the total number of ways is 5!*6*5!=86,400.

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Hope it helps.
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Manager
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Re: There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]

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29 Jan 2013, 11:08
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Cut the circle in between the two men sitting together and straighten it into a line: MFMFMFMFMFM
Now there are 6! ways to permute the men and 5! ways to permute the women.
Answer: $$6!*5!$$
_________________

Sergey Orshanskiy, Ph.D.
I tutor in NYC: http://www.wyzant.com/Tutors/NY/New-York/7948121/#ref=1RKFOZ

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Magoosh GMAT Instructor
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Re: There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]

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29 Jan 2013, 11:59
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antoxavier wrote:
There are 6 gentlemen and 5 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are seated together.
A. 85400
B. 36400
C. 86400
D. 65600
E. 88600

Arrangements in circles are tricky in that we have to be careful not to count the same arrangement multiple times. For example, with five people, the arrangements ABCDE and DEABC look different but are actually the same --- every person has the same people on each side. We have just rotated the arrangement around the circle, but we don't count that as a "new" arrangement.

One way to make sure we don't overcount is to pick one distinctive feature of the arrangement and orient everything from that. Here, we are particularly fortunate. If we had five males and five females, they could simply be alternating. The fact that we have six males and five females means there is exactly one place around the table where two guys are sitting next to each other.

Starting from the two guys sitting next to each other, we have M M F M F M F M F M F

Those six guys can be rearranged in those six slots, and no arrangement will be counted twice --- the clump of two guys breaks the symmetry, so each arrangement is unique. That's a permutation of six individuals, 6P6 = 720.
Then the females be arranged in those five slots --- again, since the two guy group breaks the symmetry, any arrangement of the women will be unique. That's 5P5 = 120.
By the Fundamental Counting Principle, we multiply these to get the total number of seating arrangements.
Total number of possibilities = (720)*(120) = 86400, answer C.

Notice, this question would become considerably more difficult if we merely changed it to seven guys and five woman --- now twelves total seats, and it's still the case that no two women can sit together.
Now, we could have a clump of three guys sitting together, and then all the other guys alternating with women ---- OR, we could have two clumps of two guys next to each other, and these two clumps could be distributed at different positions around the table --- we would have to count all those arrangements, and then for each one, figure out all the permutations of males & females. A much trickier question.

Does all this make sense?

Mike
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Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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Manager
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Re: There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]

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29 Jan 2013, 20:54
mikemcgarry wrote:
One way to make sure we don't overcount is to pick one distinctive feature of the arrangement and orient everything from that. Here, we are particularly fortunate. If we had five males and five females, they could simply be alternating. The fact that we have six males and five females means there is exactly one place around the table where two guys are sitting next to each other.

There is a more reliable way to cut the circle. Pick one particular man and call him "John". Fix his place -- say, he is the host and sits at the head of the table. Now the remaining ten people have to arrange themselves on a line; John is effectively out. We have 5! ways to order the men, 5! ways to order the women, and 6 ways to pick 5 spots out of 10 so that no two spots are adjacent. Of course, this approach with picking non-adjacent spots is difficult to generalize, but fixing one man to break the circle works quite often.
(In fact, it is not that difficult. There are $$C(n-k,k)+C(n-k,k-1)$$ ways to choose k spots out of n spots on a line so that no two chosen spots are adjacent.)

mikemcgarry wrote:
Notice, this question would become considerably more difficult if we merely changed it to seven guys and five woman --- now twelves total seats, and it's still the case that no two women can sit together.
Now, we could have a clump of three guys sitting together, and then all the other guys alternating with women ---- OR, we could have two clumps of two guys next to each other, and these two clumps could be distributed at different positions around the table --- we would have to count all those arrangements, and then for each one, figure out all the permutations of males & females. A much trickier question.

Let me suggest another solution that can be generalized. First the original problem. Let each woman pick herself a partner - a man who will be sitting on her left side. Since all women are different, we get $$6*5*4*3*2=6!$$ possibilities for forming five couples. Now we have five couples and one single man to be arranged around the circle. The single man breaks the circle, and we get $$5!$$ arrangements.

If there are 7 men and 5 women, we have $$7*6*5*4*3=7!/2$$ ways to form five couples. Now we have five couples and two single guys around the circle. Effectively, we have removed the condition of two women not being adjacent. (Now we have men and couples; before we had men and women.) One man, John, breaks the symmetry, as before. Now we have a line with one man and five couples to be arranged in some order. No need to count any combinations -- there are $$6!$$ ways to do that: both the man and the couples are unique. For 8 men and 5 women we have $$8*7*6*5*4$$ ways to form the couples. Then we fix one man and arrange the remaining two men and five couples in $$7!$$ ways. For $$M$$ men and $$F$$ women, assuming $$M>F$$, we get $$\frac{(M!)(M-1)!}{(M-F)!}$$. This also happens to work for $$M=F$$ since we can use a couple to break the circle. Lastly, if you want the minimum distance between two women to be 2, each can choose two partners -- one sitting close to her, the other a little farther away
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I tutor in NYC: http://www.wyzant.com/Tutors/NY/New-York/7948121/#ref=1RKFOZ

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Re: There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]

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27 May 2016, 22:10
antoxavier wrote:
There are 6 gentlemen and 5 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are seated together.

A. 85400
B. 36400
C. 86400
D. 65600
E. 88600

First arrange the Gents. This is a circular arrangment. 6 Gents can be arranged in 5! ways.

Now there are 6 spaces (space in between two gents) and we have to seat 5 ladies there.

First select 5 places out of 6 available places. can be done in 6C5 ways. Then arrange 5 ladies there in 5! ways.

Total: 5!*6C5*5! = 86400

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Re: There are 6 gentlemen and 5 ladies to dine at a round table. [#permalink]

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05 Jun 2017, 04:15
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Re: There are 6 gentlemen and 5 ladies to dine at a round table.   [#permalink] 05 Jun 2017, 04:15
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