2018 AMC 10A Problem 4

Split the job into two questions: which periods get math, and which course goes in each.

1. Which periods? We need three of the periods $1$ to $6$ with no two next to each other. List them in order, smallest first.
- Start with $1$: the next can be $3$, then $5$ or $6$. Or the next can be $4$, then $6$. That gives $\{1,3,5\}$, $\{1,3,6\}$, $\{1,4,6\}$.
- Start with $2$: the only choice is $\{2,4,6\}$.
- Start with $3$ or later: the next two would need periods $5$ and $7$, but there is no period $7$. Nothing works.

So there are $4$ sets of periods.
2. Which course goes where? In a chosen set, the first math period can hold any of the $3$ courses, the second any of the $2$ left, and the third the last one. That is $3 \cdot 2 \cdot 1 = 6$ orders.
3. Put it together. Each of the $4$ sets can be filled in $6$ ways: $4 \cdot 6 = 24$.

The answer is $\boxed{\textbf{(E)}\ 24}$.

The "not next to each other" rule only cares about the positions, not about which course is where. So handle the positions first, where the rule lives, and then arrange the courses freely. This two-step split ("where" and then "what") works for many seating and scheduling problems.

Count the bad position sets and subtract. There are $\binom{6}{3} = 20$ ways to pick three periods with no rule. List the sets that include at least one pair of neighbors: there are $16$ of them (for example $\{1,2,4\}$, $\{3,4,5\}$), so $20 - 16 = 4$ good sets remain, and $4 \cdot 6 = 24$ as before. Listing the $4$ good sets directly is quicker.

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 $\boxed{\textbf{(E)}\ 24}$.