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.

• The trick to the problem is determining how many ways the two friends can sit together. – Chris K Dec 1 '13 at 0:49

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