[Math] 3 Gents, 3 Ladies, and one round table

Question

The question is how many ways the aforesaid ladies and gentlemen be seated around the round table, such that two and only two ladies sit together.

The answer says 72.

My understanding:

(#ways to arrange 3 gentlemen at the table)                     // (3-1)!
*(#ways to select 2 out of 3 ladies)                            // 3P2
*(#ways to arrange the pair of ladies among themselves)         // 2

= 2*6*2 = 24

So where am i going wrong ?

EDIT :

(#ways to arrange 3 gentlemen at the table)                             // (3-1)!
*(#ways to select 2 out of 3 ladies)                                    // 3P2
*(#ways to select 2  out of 3 available positions in between the men)   // 3P2

= 2*6*6 = 72

Am i understanding correctly ?

Answer 1

There are $\binom{3}{2}$ to select the two women who sit together and $2!$ ways to arrange them within that block. That leaves four open seats, two of which are adjacent to the two women who sit together, so there are $2$ ways to sit the third woman in a seat that is not adjacent to the other two women and $3!$ ways to arrange the men in the remaining seats. Thus, there are $$\binom{3}{2} \cdot 2! \cdot 2 \cdot 3! = 3 \cdot 2 \cdot 2 \cdot 6 = 72$$ ways to seat three men and three women at a round table if exactly two of the women sit together.

Seating the men first is more difficult. You have to choose which two men are to sit three seats apart (to accommodate the pair of women who sit together), arrange those two men in those seats, then choose which of the two remaining seats will be filled by the third man. This can be done in $\binom{3}{2} \cdot 2! \cdot 2$ ways. You can then choose two of the three women to sit together, arrange them in two ways, and seat the third woman in the remaining seat, which can be done in $\binom{3}{2} \cdot 2! \cdot 1$ ways. Thus, the number of possible arrangements is $$\binom{3}{2} \cdot 2! \cdot 2 \cdot \binom{3}{2} \cdot 2! \cdot 1 = 3 \cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 1 = 72$$