# Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together.

Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together.

My attempt:

6 people may be 3 men and 3 women, 2 men and 4 women or 1 man and 5 women. Then by the gap method, 3 men and 3 women can be seated in $\binom{5}{3}2!3!=120$ ways.
2 men and 4 women can be seated in $\binom{5}{4}3!\frac{4!}{2!}=360$ ways
1 men and 5 women can be seated in $\binom{5}{5}4!\frac{5!}{4!}=120$ ways

So the total number of ways is 600, but the answer given in my book is 5400. Where did I go wrong?

• Are you asking whether the question means exactly one of the two men is seated or at most one of the two men is seated? Jan 7, 2015 at 11:39
• My interpretation is that no two men can sit next to each other at the table. Then you would need at least 3 women to keep the men separated. Jan 8, 2015 at 3:34

Label the chairs from $1$ to $6$ in increasing order, label the men and the women from $1$ to $5$. Consider three cases:

1) There is 1 man: choose a man in $5$ ways, place the man at chair $1$, and permute the $5$ women in the remaining chairs in $5!$ ways. Therefore the number of ways is $5\cdot 5!=600$.

2) There are 2 men: choose two men in $\binom{5}{2}$ ways, place the man with the smallest label at chair 1 and the other at chair $3$, $4$ or $5$ ($3$ ways), choose four women in $\binom{5}{4}$ ways and arrange them in the remaining chairs in $4!$ ways. Therefore the number of ways is $\binom{5}{2}\cdot 3\cdot \binom{5}{4}\cdot 4!=3600$.

3) There are 3 men: choose three men in $\binom{5}{3}$ ways, place the man with the smallest label at chair 1, and the other at chair $3$ and $5$ ($2$ ways), choose three women in $\binom{5}{3}$ ways and arrange them in the remaining chairs in $3!$ ways. Therefore the number of ways is $\binom{5}{3}\cdot2 \cdot \binom{5}{3}\cdot 3!=1200$.

Hence the total number of ways is $600+3600+1200=5400$.

You forgot to include the number of ways in which the men can be selected. In this case, you get:

$${5 \choose 3}{5 \choose 3}2!3! = 1200$$ $${5 \choose 2}{5 \choose 4}3!\frac{4!}{2!} = 3600$$ $${5 \choose 1}{5 \choose 5}4!\frac{5!}{4!} = 600$$

This results in a total of $1200 + 3600 + 600 = 5400$ possibilities.

My interpretation is that 2 specific men are never together. In that case, my solution would be

Choosing 6 out of 8 persons(when those 2 are not chosen) + choosing 4 out of 8 persons( 2 men are already chosen) .

(8,6).5! + (8,4).3!.(4,2).2!

And if question is interpreted as any of 2 men can never be together, then you will need at least 3 women. In that case, solution would be Summation of cases choosing 3,4,5 women.

(5,3).2!.(5,3).(3,3)3! + (5,4).3!.(5,2).(4,2).2! + (5,5).4!.(5,1).(5,1).1!

The question is incomplete. The question does not specify if the seats are labeled, such that a cyclic shift of the people sitting on the table counts as a different seating. If the seats are not labeled, a cyclic shift would result in the same seating as before the cyclic shift. So without labels on each seat of the round table,

$m_1 f_1 m_2 f_2 m_3 f_3 == f_1 m_2 f_2 m_3 f_3 m_1$

but with labels that's no longer true because we can see that on the left hand side the first seat is taken by m_1, whereas it is taken by f_1 on the right hand side. (Assuming we chose men 1 to 3 and women 1 to 3).

I would argue that the seats are labeled, because in the real world the table is located in a room where seats can be labeled with the position in the room.

For the case of 3 men, if the seats are labeled, there are

$${5 \choose 3}{5 \choose 3}6 * 3 * 2 * 2 = 7200$$

possibilities, but if they are not labeled, there are

$${5 \choose 3}{5 \choose 3}2!3! = 1200$$

This result is in agreement with by the fact that there are 6 possible shifts on a table of 6 and 7200/1200 = 6.