A class contains $5$ boys and $5$ girls.What is the probability that some two girls will sit next to one another? A class contains $5$ boys and $5$ girls.For the class banquet,they select seats at random around a circular table that seats $10$. What is the probability that some two girls will sit next to one another ?
I've reasoned as follows about the problem:
Let $A$ be the even that some two girls sit next to each other.
Then $P(A)=1-P(\lnot A) $
So the problem now breaks down to find all the combinations in which no two girls sit next to one another.
For that I have considered to let the $5$ girls sit around the table ,without the boys.
I can have $4!$ distinct combinations,and the same goes if I would let the $5$ boys sit around the table without the girls.
Now my intuition here was that whatever combination I have of the $5$ girls around the table,I can always "overlap" the combinations of the $5$ boys in between each girl .
So that I have $4!\cdot 4!$ combinations in which no girl sit next to another girl,which yields a probability $P(\lnot A)=\cfrac{4!4!}{9!}$
Finally I have that $P(A)=1-P(\lnot A)=1-\cfrac{4!4!}{9!}\approx 99 \% $
This last result really leaves me a bit skeptical,I belive I have done some mistake conceptually...
 A: Forget the circle for a minute.
The total number of permutations is $10!=3628800$.
The number of permutations with no two girls sitting next to each other is $5!\cdot5!\cdot2=28800$.
Hence the probability of two girls sitting next to each other is $1-\frac{28800}{3628800}\approx99.2\%$.
A: Suppose that one of the girls is Abigail.  Seat her first.  As we move counterclockwise around the table starting with the seat to her immediate right, we can seat the remaining $9$ people in $9!$ ways.
If no two girls sit next to each other, then the boys and girls must sit in alternating seats, so there are four seats available to the remaining girls once Abigail has taken her seat.  We can arrange the other four girls in those seats in $4!$ ways as we move counterclockwise around the table from Abigail.  This leaves five seats for the boys.  Relative to Abigail, they can be arranged in these seats in $5!$ ways as we move counterclockwise around the circle from the seat to her immediate right.  Hence, the number of arrangements in which no two girls sit together is $4!5!$.  
Hence, the probability that two of the girls sit together is 
$$1 - \frac{4!5!}{9!}$$
A: It looks like there are a few correct answers, but I think there is a significantly easier way to do this. First, we don't need to distinguish between different girls or different boys, so instead let's approach the problem in terms of $5$ red balls and $5$ blue balls. Second, a circular table can be thought of as a rectangular table, we just have to be careful about the two end seats.
Now, the question is how many ways can we line up $5$ red balls and $5$ blue balls such that no two red balls are next to eachother or on opposite ends of the line. Choosing to place a ball in position 1 forces everything else, so there are only $2$ ways(either red first or blue first). How many ways all together are there to line up the balls? $\binom{10}5$ giving us $$P=1-\frac{2}{\binom{10}5}\approx 99.2\%$$
