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

Bunuel and SergeyOrshanskiy already gave very clear answers, but because antoxavier originally asked about this through Magoosh, I want to explain a little further.

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

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.)


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

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

C is the answer.

To only count relative positioning, we divide the # of possible gents permutations by the number of gents. This yields 5! instead of 6!.

However, why does he not make sense to do 6!*# of possible women permutations, and then divide that total amount by the number of seats on the table?

To me, it makes more sense to divide in the end by the number of seats. Indeed, at the end you could rotate everyone by 1 or 2 or n seats and still get same relative positioning. Why is this incorect?

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.

Desired arrangement: MFMFMFMFMFM

Step 1:
Number of arrangements for 6 Men at a round table = 6! = 720

Step 2:
Number of ways to arrange women such that no 2 women sits next to each other = 5! = 120

Total possible arrangements = 720*120 = 86400

Ans: C

Hi, can you please explain, how did we come to reach "6 men can be arranged around a table in (6-1)!=5! ways"? And also how are 6 spots available between the men?

Hi

What if the chairs were fixed. In this case how do we proceed?
Lets say if we assign seat No.1 to Male 1. Seat No.3 we have 5 ways of being occupied because there are 5 other men (M2,M3,M4,M5 & M6). Seat No. 5 can be occupied in 4 ways by 4 men ( M3,M4,M5& M6) and so on giving us 5! ways of arranging the men( Seat No 11 & 1 which are adjacent as this is a circular arrangement are occupied by men). This leaves us with even seats (Seat 2,4,6,8,10). 5 women will occupy these 5 seats in 5! ways. Shouldn't the answer be 5!*5!?
Although I do understand your method I just want to know what is the flaw in my logic ?

Thanks

@bunnel

for men = 5! ways
now we have 6c5 ways to make women sit ; and for women 5! ways
5!*6c5*5! = 86400
IMO C

Hello Bunuel,

Can we solve this question if we seat ladies first? Someone I started solving this question by adjusting ladies first and then men.

I have a question:
Let's see the arrangement in a non-circular way:

M F M F M F M F M F M
now, men can be arranged in 6! ways and females can be arranged in 5! ways

But when we make it circular, then,
M1 F1 M2 F2 M3 F3 M4 F4 M5 F5 M6 and
M6 F5 M5 F4 M4 F3 M3 F2 M2 F1 M1 are same

So, we divide it by 2
total number of combinations = (6! * 5!) / 2

There is no difference in the first arrangement and second arrangement if you place them at a round table. Hence, you can't divide it by 2.

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