3 Gents, 3 Ladies, and one round table 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 ?
 A: 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 
$$\binom{3}{2} \cdot 2! \cdot 2 \cdot \binom{3}{2} \cdot 2! \cdot 1 = 3 \cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 1 = 72$$
A: There are $3$ ways to select the lady that is to be alone, and two ways to seat the ladies (select which of the two is going to be to the left). After this the problem breaks down to seating three men and a woman in a line so that the lady is not at one of the ends. There are $2$ possible seats for the lady and after that $3!$ ways to seat the men. So all in all there are $3\cdot2\cdot 2\cdot 3!=72$ ways to do it.
Note that for the answer to make sense reflecting a solution gives a new solution, but rotating doesn't.
A: 3c2*(5-1)!=72
Any two ladies sit together=3C2
Remaining positions=(5-1)=4!
Total ways=3C2*4!=72
A: First of all we select 2 ladies out of three and permute them , (3C2 x 2!) 
Now as two ladies are seated around the table, We left with 3 gents and 1 lady and this lady is not allowed to sit adjacent to both the ladies who seated already. So, this lady has 2  options out of 4, where she will be seated, therefore select that place and permute ( 2C1 x 1!) 
Now all the 3 ladies are seated, then we left with 3 places and we have 3 gents which can be arrange in 3! Way. 
Therefore our answer is 
(3C2 x 2!) x (2C1 x 1!) x 3! = 72 
