Find the number of ways in which 6 gentlemen and 3 ladies
can be seated around a table so that every gentleman may have a lady
by his side.
Consider, GLGGLGGLG (G-> Gentleman L-> Lady)
1)The gentlemen can be seated in $6!$ ways.
2)The 3 ladies can sit in any of the 3 spots between the gentlemen in
$3!$ ways.
3)The ladies and gentlemen can be switched.
4)They can be rotated in 9 ways which are all equivalent.
Hence, $\frac{(6!)(3!)(2)}{9}$
Where have I committed my mistake?
I checked this – Similar question But I want to find where I am going wrong.
I know that we can do it by arranging the gentlemen in $5!$ ways and the ladies can be arranged in 3! ways. Both can be switched in $2!$ ways.
Hence, $5!3!2!$
But, I want to do it in a method where I find the number of ways in which the gentlemen can be arranged in rows. Then remove the equivalent arrangements from that.
Best Answer
Method 1: Suppose Adam is one of the men. We will use him as our reference point.
Seat Adam at the table. Relative to him, the remaining men can be seated in $5!$ ways as we proceed clockwise around the table. Since there must be a woman next to Adam and there must be two men between each pair of women, there is either a woman to Adam's left or to his right, but not both. Making this choice determines the positions of the women.
The women can fill those three positions in $3!$ ways. Hence, there are $5!3!2!$ ways to arrange six men and three women at a round table in which each man sits next to a woman, as you found.
Method 2: We form a linear arrangement with a man in the first position, then join the ends of the row to form an admissible circular arrangement.
We can arrange the six men in $6!$ ways. If we leave a man in the leftmost position, we must insert the first woman next to him, which yields an arrangement of the form $$GLGGLGGLG$$ or next to the man in the second position, which yields an arrangement of the form $$GGLGGLGGL$$ We can arrange the three women in each of the two arrangements in $3!$ ways. We then join the ends of the row to form a circular arrangement. Hence, there are $6!2!3!$ linear arrangements with a man in the first position that correspond to an admissible circular arrangement.
If we now seat the people at the table so that we proceed clockwise from the first man in the linear arrangement, the first arrangement corresponds to the diagram at left above while the second arrangement corresponds to the diagram at right above (where the first man in the linear arrangement occupies the role that Adam played above). Since there are six men who we could use as our reference point for the same circular arrangement, there are $$\frac{6!2!3!}{6} = 5!2!3!$$ distinguishable ways to arrange six men and three women at a round table so that each man sits next to a woman.
Notice that the man we choose as our reference point acts as a reference point for both the men and the women.
Method 3: Suppose Amanda is one of the women. We will use her as our reference point.
Seat Amanda at the table. Relative to her, we can seat the remaining two women in $2!$ ways as we proceed clockwise around the table.
Since each man must sit next to a woman, there must be two men between each pair of women. Relative to Amanda, the men can be seated in $6!$ ways as we proceed clockwise around the table.
Hence, there are $2!6!$ seating arrangements of six men and three women at a round table in which each man sits next to a woman.
Method 4: We form a linear arrangement with a woman in the first position, then join the ends of the row to form an admissible circular arrangement.
Suppose we place a woman in the first position. Then the remaining two women must be in the fourth and seventh positions to ensure that each man sits next to a woman when we join the ends of the row to form a circle. The women can be arranged in $3!$ ways and the men can be arranged in $6!$ ways within the row. Hence, there are $3!6!$ ways to form linear arrangements with a woman in the first position that correspond to an admissible circular arrangement.
We join the ends of the row to form a circular arrangement, proceeding clockwise around the table from the woman in the first position. Since there are three women who we could use as our reference point for the same circular arrangement, there are $$\frac{3!6!}{3} = 2!6!$$ distinguishable circular arrangements of three women and six men are seated at a round table in which each man sits next to a woman.
Notice that the woman we use as our reference point serves as a reference point for both the women and the men.