Find the number of eight-letter words that use letters from the set $\{A, B, C\}$ and contain exactly three A's.
There are $\binom{8}{3}$ ways to choose the positions for the A's. Each of the five remaining positions can be filled with a B or a C. Hence, the number of such words is
$$\binom{8}{3}\cdot 2^5$$
How many of these words contain no two consecutive A's?
Consider a five-letter word composed of B's and C's such as $$BBCBC$$ We wish to insert three A's so that no two of them are consecutive. We have six spaces in which to insert the three A's, indicated by the squares in the example.
$$\square B \square B \square C \square B \square C \square$$
To ensure that no two A's are consecutive, we must choose three of these six spaces into which to insert an A. For instance, if we choose the first, second, and fifth spaces, we obtain the sequence
$$ABABCBAC$$
There are $2^5$ five-letter words composed of B's and C's and $\binom{6}{3}$ ways to choose three of the six spaces in which to insert an A. Hence, there are $$2^5 \cdot \binom{6}{3}$$ eight-letter words composed from the letters A, B, and C with exactly three A's in which no two of the A's are consecutive.