How many ways can a television producer air commercials... A television director has to schedule commercials during 6 time slots.
There are four distinct commercials, A, B, C, and D. A must be aired three times but never twice consecutively.
How many ways are there to air the ads?
 A: Let $X,Y$, and $Z$ stand for the commercials $B,C$, and $D$ in any order. Counting the first and last positions, there are four possible slots for $A$, indicated by underscores: $\_X\_Y\_Z\_$. We can put at most one $A$ commercial in each of those slots, so to schedule the $A$ commercials, we need only decide which $3$ of the $4$ possible slots to use; we can do that in $\binom43$ ways. Then we have to decide on the order of $B,C$, and $D$ in the $X,Y$, and $Z$ slots. How many permutations of $A,B$, and $C$ are possible? Can you finish it from there?
A: There is some ambiguity in the problem: must each of $B$, $C$, and $D$ be shown? 
Yes, they all must be shown: Call the time slots $1, 2, 3, 4, 5, 6$. Then $A$ is either in $1,3,5$ or in $2,4,6$, two possibilities. For each of these, there are $3$ empty slots, which can be filled in $3!$ orders from $B$, $C$, and $D$, for a total of $(2)(3!)$.  This is likely to be the intended interpretation.
No, they need not be all show: this is a much less plausible interpretation. The location of the $A$'s can be chosen in $2$ ways. For each of these, the first empty slot can be filled in $3$ ways. For each such possibility, the second empty slot can be filled in $3$ ways, and then the third can be filled in $3$ ways, for a total of $(2)(3^3)$.
