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I am looking at the following exercise:

$5$ women sit at a round table. $3$ men come later. With how many ways can the men sit between the women, if no man can sit next to an other man?

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I have done the following:

We have the following:

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right?

The first man has $5$ possibilities to sit between two women. The second man has $4$ possibilities to sit between two women. The third man has $3$ possibilities to sit between two women.

Therefore, we get that there are $5\cdot 4\cdot 3$ ways so that the men sit between the women, if no man can sit next to an other man.

Is this correct?

If yes, is the justifications correct and complete? Could we improve that or make that more formally?

Is it important that the table is round or isn't there a difference if it would have an other shape?

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Your method is correct. Alternatively we can do this as follows:

We have to choose three places from five. This can be done in $\binom{5}{3}$ ways. Now these men can be arranged in $3!$ ways. Thus the answer is $\binom{5}{3} 3!$

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  • $\begingroup$ We don't have to consider how the women sit, do we? Is it important that the table is round or isn't there a difference if it would have an other shape? $\endgroup$ – Mary Star Nov 6 '17 at 13:59
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    $\begingroup$ None at all. No difference. And no we don't have to consider the women, because they are already in their seats. $\endgroup$ – samjoe Nov 6 '17 at 14:02
  • $\begingroup$ As I did it in my post and when we multiply all the arrangements of men, we get $5\cdot 4\cdot 3\cdot 3!$. Your result is $\binom{5}{3}\cdot 3!$. It is different. Where is my mistake? $\endgroup$ – Mary Star Nov 6 '17 at 14:07
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    $\begingroup$ You don't need to arrange the men in your method! Its already fine. You have already considered them if you do it in your way. $\endgroup$ – samjoe Nov 6 '17 at 14:08
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    $\begingroup$ In your method, you chose seat, and put the man there. You take care of all possibilities. First there are 5 ways, then 4, then 3. There is no other way for them to sit. In my method, you choose the three seats together, and then put men there. By $\binom{5}{3}$, we only chose $3$ seats. This number corresponds to only choosing the seats. Arrangement is not considered. So we multiply by $3!$. $\endgroup$ – samjoe Nov 6 '17 at 14:20
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Let us first arrange the 5 women around the table. Now, this may seem to be 5!, but it is wrong since the order in which they are seated is irrelevant i.e.,

consider this order round the table {W1 , W2 ,W3, W4, W5} in seats { S1, S2, S3, S4, S5}.

Now, even if you shift each women by a seat to left/right, their neighbours will remain the same i.e., {W2 ,W3, W4, W5, W1} in seats { S1, S2, S3, S4, S5}, the combination will still remain the same. This is unlike the linear arrangement where this is considered as a new arrangement.

So, if you calculate this the no. of ways of seating n people round the table when the order matters is (n-1)!. So, (5-1)! ways are possible.

Now, there are 5 spots left for 3 men to choose from. So, allocating 3 spots among 5 in 5P3 ways.

So, total no. of ways are 4! * 5P3 ways

=> (5!/2!)*4! ways are possible.

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  • $\begingroup$ It doesn't really matter what's the shape of the table unless, you consider a member sitting at different sides of the table (with their neighbours being the same) as a new combination. $\endgroup$ – Your IDE Nov 6 '17 at 14:15
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Since this is abround table problem, let's fix a point. Let's fix woman A on the first position. We now have 4+3 = 7 positions to decide

We now have to choose 3 out of those positions in a way that none of those are next to each other. We can do this by hand: consider the positions 1 through 8, we have the following possibilities: (1,3,5), (1,3,6),(1,3,7),(1,4,6),(1,4,7),(1,5,7),(2,4,6),(2,4,7),(2,5,7),(3,5,7)= 10 possibilities

So now you have to decide which of the men gets which of these 3 seats and which of the women gets which of the 4 seats (the first woman we don't count because we "fixed" her position)

For the men, we have 3! Choices, for the women we have 4!

So the answer is 10*3!*4!

Edit: to explain the round table thingy

In combinatorics, round tables are usually regarded in the same way. Let's say you have people A,B,C,D sit at a round table. The number of different ways they could sit down is 4!, Same as the permutations of the word ABCD. But when we talk about a round table, we usually consider the cases ABCD, BCDA, CDAB and DABC to be equal (because they'reall in the same order respect to each other). That means every case is being counted 4 times. So we divide 4! by 4 and we get 3!. This is the same as "fixing" the position of one of them and ignoring them entirely. If you ignore one of the 4 people and organize the rest, we get 3! ways of ordering 3 people

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  • $\begingroup$ Is it important that the table is round or isn't there a difference if it would have an other shape? We have that the women already sit at the table when the men come. Do we have to consider also the way that the women sit? $\endgroup$ – Mary Star Nov 6 '17 at 14:01
  • $\begingroup$ Edited for the round table explanation $\endgroup$ – Francisco José Letterio Nov 6 '17 at 14:13

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