5 friends are sitting together, what is the probability that 2 are NOT sitting together five friends including Bilyana and Bojana are sitting in a row in a theatre determine the probability that they are not sitting together.
This is part of a homework assignment. I don't even know where to start with this.
 A: We will put on a couple of the seats a sign saying "Only to be used by Bilyana or by Bojana. Anyone else, Bulgarian or not, keep off."
There are $\binom{5}{2}$ equally likely ways to select the two seats. Exactly $4$ of these choices leave us with the two B's sitting together. Thus our probability is $1-\frac{4}{\binom{5}{2}}$. 
A: Hint:
Think of sitting Bilyana first, Bojana second and the rest next. How many ways of choosing a seat is there for Bilyana? and then for Bojana? (consider there's a different answer to this if Bilyana is (is not) seating at one end) What about the others next?
Work on this and if you still can't find your answer, update the question including your work so far.
A: First you need to figure out the sample space. 
Since there are five friends and 5 seats in a row, you need to line them up. 
5 x 4 x 3 x 2 x 1 = 120
Next you need to find how many ways they can sit together. 
2 x 1 x 3 x 2 x 1 = 12 
Since there are 4 other cases of them sitting together multiply them. 
12 x 4 cases= 48
The probability of them sitting together is 48/120
Since this is the probability of them sitting together, subtract 120-48= 72
This is assuming that the other cases are ones where they aren’t sitting together. 
So it’s 72/120 or 60%
