Find the number of ways in which $6$ men and $5$ women can dine at a round table if no two women sit together. Find the number of ways in which $6$ men and $5$ women can dine at a round table if no two women sit together. 
Now men can sit in $6!$ ways and women can sit in $P^{6}_{5}$ ways, but the textbook says a different answer. Can you explain?
 A: Initially anybody can seat anywhere on circle. So we fix a person to apply basic principle of multiplication. 
To understand this how is it so,
Consider two circular arrangement of 5 persons A,B,C,D,E we start from A and hence end with one arrangement A,B,C,D,E If You begin with B rather and End with a one arrangement B,C,D,E,F,A 
In a circle this both arrangement are same due yo cyclic nature that was not so in arragement in row.
So to avoid repetition we initially fix one person for starting point.
Now your answer  there are no restrictions on men so we can arrange them in 5! Ways. Now we are left with 6 places between men where we can place women so that no two sit together. Now selection of sit and arrangement can be done in 6P5 ways. So total 5!×6P5 that's 6!5! Ways.
A: Two circular seating arrangements are considered to be distinguishable if the relative order of the people differs.  

In how many ways can six men sit at a round table?

Method 1:  Consider the diagram below.

We can describe a particular seating arrangement by describing the order of the men as we proceed counterclockwise (anticlockwise) around the circle.  Notice, however, that we can describe this particular seating arrangement in six ways, depending on whether we start with A, B, C, D, E, or F.  Thus, this particular circular arrangement corresponds the six linear arrangements ABCDEF, BCDEFA, CDEFAB, DEFABC, EFABCD, FEABCD.  More generally, each circular arrangement of the six men corresponds to six linear arrangements, corresponding to the six possible starting points as we proceed counterclockwise around the circle.  Since six men can be arranged in a row in $6!$ ways, there are 
$$\frac{6!}{6} = 5!$$
distinguishable circular arrangements of the men.  
More generally, we can arrange $n \geq 1$ distinct objects in a circle in 
$$\frac{n!}{n} = (n - 1)!$$
ways since each circular arrangement corresponds to $n$ linear arrangements, one for which each of the $n$ starting points as we proceed counterclockwise around the circle.
Method 2:  We consider arrangements relative to the first man we sit at the table.   
We seat man A.  We use him as our reference point.  As we proceed counterclockwise around the circle, we can seat the remaining five men in $5!$ orders relative to man A.
More generally, given a set of $n$ distinct objects, we can arrange them in $(n - 1)!$ orders as we proceed counterclockwise around the circle relative to the first object we place on the circle.

In how many ways can six men and five women dine at a round table if no two women sit together.

We can seat the men in $5!$ distinguishable ways.  This create six spaces, one to the immediate right of each man.  To ensure that no two of the women sit in adjacent seats, we must choose five of those six spaces in which to insert a woman.  Once those spaces have been selected, we can arrange the five women in those spaces in $5!$ orders.  Hence, the number of possible seating arrangements is 
$$5! \binom{6}{5} \cdot 5! = 5!P(6, 5)$$  
