Three non-adjacent A's

Text

In how many ways can $3$ letters A and $7$ letters B be arranged in a row so that no two A’s are ever adjacent?

Solution

First place the $7$ B’s in a row: $\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_$ There remain $8$ “gaps” (the $6$ internal ones plus the $2$ external ones) in which to insert the A’s. Since no two A’s may be adjacent, each gap holds at most one A: it is enough to choose $3$ of the $8$ gaps. The B’s are all identical and so are the A’s, hence $N = \binom{8}{3} = \frac{8\cdot 7\cdot 6}{6} = \boxed{56}.$