In how many ways can $3$ ladies and $3$ gentlemen be seated around a round table so that any two and only two of the ladies sit together?

Question

Q) In how many ways can $3$ ladies and $3$ gentlemen be seated around a round table so that any two and only two of the ladies sit together?

Answer given- $72$

My answer - $108$

My Approach-

Number of required ways= total number of ways of seating two women adjacent - number Of ways of seating $3$ women adjacent.

$3C2 \cdot 2 \cdot 4!-3C3 \cdot 3! \cdot 3! =144-36=108$

Where am I going wrong?

Answer 1

HINT:

Try thinking about this as follows

Two woman will sit next to eachother: $$ WW $$ this can be done in $\binom{3}{2}\cdot 2$ different ways. Next to them on both sides we will need men $$ MWWM $$ this can again be done in $\binom{3}{2}\cdot 2$ different ways.

Can you figure the rest out?

Answer 2

Method 1: We subtract the number of arrangements in which no two women sit together and in which all three women sit together from the total number of arrangements of six people at a round table.

Suppose Alexander is one of the men. Seat him. Since only the relative order of the people matters, it does not matter where Alexander sits. The other five people can be seated in $5!$ orders as we proceed clockwise around the table from Andrew.

If none of the women sit together, the men and women must alternate. Again, we seat Alexander first. Seating him determine which seats will be occupied by the men and which will be occupied by the women. The other two men can be seated in $2!$ ways as we proceed clockwise around the table from Alexander. The three women can be seated in $3!$ ways as we proceed clockwise around the table from Alexander. Thus, there are $2!3!$ seating arrangements in which no two women sit together.

For arrangements in which all three women sit together, we again seat Alexander first. We have three objects to arrange relative to Alexander, the other two men and the block of three women. The objects can be arranged in $3!$ ways as we proceed clockwise around the table relative to Alexander. The women within the block can be arranged in $3!$ orders. Hence, there are $3!3!$ ways the men and women can be seated if all the women sit together.

Thus, there are $5! - 2!3! - 3!3!$ admissible arrangements.

Method 2: We correct your count.

We seat Alexander first. We have four objects to arrange relative to Alexander, the block of two women and the other three people. Choose which two of the three women form the block, which can be done in $\binom{3}{2}$ ways. The four objects can be arranged in $4!$ ways as we proceed clockwise around the table from Alexander. The two women in the block can be arranged in $2!$ ways. Hence, there are $\binom{3}{2}4!2!$ ways in which at least two women sit together, as you correctly calculated.

However, we have counted each arrangement in which all three women sit together twice, once for the first two women we encounter as we proceed clockwise around the table from Alexander and once when we encounter the last two women as we proceed clockwise around the table from Andrew. Since we want the number of arrangements in which exactly two women sit together, we do not want to count these arrangements at all. Therefore, we must subtract twice the number of arrangements in which all three women sit together.
As shown above and as you correctly calculated, there are $3!3!$ arrangements in which all three women sit together. Hence, there are $$\binom{3}{2}4!2! - 2 \cdot 3!3!$$ admissible seating arrangements.

Method 3: We do a direct count.

Seat Alexander. We then ask the women to wait while we seat the other two men. Relative to Alexander, they can be seated in $2!$ ways as we proceed clockwise around the circle relative to Alexander. This creates three spaces, one to the left of each man, in which the women can be seated. Choose which two of the three women sit together, which can be done in $\binom{3}{2}$ ways. Choose in which of the three spaces these women sit, which can be done in $\binom{3}{1}$ ways. Arrange the two women in that space, which can be done in $2!$ ways. Choose in which of the two remaining spaces between two men the other woman sits, which can be done in $\binom{2}{1}$. Therefore, there are $$2!\binom{3}{2}\binom{3}{1}2!\binom{2}{1}$$ admissible seating arrangements.