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.
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 $1-\frac{4}{\binom{5}{2}}$.
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%