2018 AMC 10A Problem 4

First decide which periods hold math, then decide which course goes in each.

Three periods out of $1, 2, \dots, 6$ with no two consecutive: list them. Each chosen period after the first must skip at least one period, so the possibilities are
$$\{1,3,5\},\ \{1,3,6\},\ \{1,4,6\},\ \{2,4,6\}.$$
That is $4$ sets of periods.

For each set, the three distinct courses can be assigned to the three periods in $3! = 6$ ways.

Total: $4 \cdot 6 = 24$. The answer is $\boxed{\textbf{(E)}\ 24}$.

Separating "where" from "what" is the standard move for restricted arrangements: the adjacency restriction lives entirely in the choice of positions, and once positions are fixed the distinct courses are freely permuted. In general the number of ways to pick $k$ non-adjacent positions from $n$ is $\binom{n-k+1}{k}$, here $\binom{4}{3} = 4$.

Complementary counting: all $\binom{6}{3} = 20$ position sets minus those with an adjacent pair. Sets with at least one adjacent pair: count pairs of adjacent periods ($5$) times a third period ($4$), which is $20$, minus double counts of three consecutive periods ($4$), giving $16$. So $20 - 16 = 4$ good sets, then multiply by $6$.

• Counting the four position sets and forgetting to multiply by 3! for which course goes where, giving 4 or 12.
• Missing $\{1,3,6\}$ or $\{1,4,6\}$ when listing, since gaps can be larger than one period.