Circular Arrangement of students $6$ male students and $3$ female students sit around a round table. The probability that no $2$ female students sit beside each other can be expressed as $\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is the value of $a+b$?
 A: One of the chairs is a throne, and the Queen, who is one of the women, sits there.
That leaves $8$ chairs, of which we must choose $2$ for the remaining women. There are $\dbinom{8}{2}$ equally likely ways to do this.
Now we count the good choices, where no two women are neighbours.The two chairs next to the Queen can't be used. So effectively we want to choose $2$ from the remaining $6$. But $5$ of these pairs consist of neighbouring chairs. So our probability is
$$\frac{\binom{6}{2}-5}{\binom{8}{2}}.$$
A: At first make the female students sit around the round table.This can be done in $(3-1)!=2$ different ways.
Now in between the female students there are $3$ places. You have to divide the $6$ male students into 3 groups. This can be done in this way: You first consider all the male students are same which implies that they can be divided into 3 groups in $\binom{5}{2}$ ways. In each such selection we can get $6!$ different arrangements. So the total number of arrangements is $10 \cdot6!$
So the total no. of ways in which the students can be seated with no two female students side by side=$2\cdot10\cdot6!$.
And total no. ways in which the students can be seated is $(9-1)!$ ways.
So the probability is $\frac{2\cdot10\cdot6!}{8!}=\frac{2\cdot10}{7\cdot8}=\frac{5}{14}$
So $a+b=19$  
A: Seat the first woman first-she can go anywhere.  The next one can go in a seat with one blank, probability $\frac 28$ or in a seat at least two away, probability $\frac 48$.  If the second is one away, the last one has four seats that are not next to one of the others out of seven.  If the second is at least two away, the third has three choices out of seven.  So the total chance is $\frac 14 \cdot \frac 47 + \frac 12 \cdot \frac 37=\frac 5{14}$
