In how many ways can an engineering student select and schedule three technical electives in his final four semesters? I've the following question:
An engineer needs to take exactly three technical electives sometime during
his final four semesters. The three are to be selected from a list often. In how many ways can he schedule those classes, assuming that he never wants to take more than one technical elective in any given term?
I tried to solve it in the following way:
Total number of ways $3$ electives can be chosen from a list of $10$ electives: $10P3$
Total number of was $3$ electives can be arranged in $4$ semesters: $4P3$
So, total number of permutation= $4P3 \cdot 10P3 = 24 \cdot 720 = 17,280$.
Is my calculation correct?
 A: Your answer is too large.  The order in which the engineering student selects the electives does not matter, just the order in which he takes those electives.  There are $\binom{10}{3}$ ways for the student to select which three electives to take and $P(4, 3)$ ways for the student to schedule the three selected courses in three of the last four semesters.  Therefore, there are $$\binom{10}{3}P(4, 3)$$ ways for the engineering student to select and schedule three technical electives in his final four semesters.
A: Just another perspective.

*

*Let $s_1s_2s_3s_4$ be the last 4 semesters arranged on a line and immobile. Make 3 identical cards, each with label $T$  and 1 card with label $F$. There are $\frac{4!}{3!\times1!}$ ways to distribute these 4 cards. A permutation $TFTT$, for example, means that we have no technical elective taken in semester $s_2$.


*Let $e_1e_2\ldots e_9e_{10}$ be the technical electives arranged on a line and immobile. Let's take a permutation $TFTT$ as an illustration. Make 7 identical cards, each with label $s_2$ (we only duplicate the semester associated with $F$), 1 card with label $s_1$, 1 card with label $s_3$, and 1 card with label $s_4$. There are $\frac{10!}{7!\times1!\times1!\times1!}$ ways to distribute these 10 cards to technical electives.
As there are $\frac{10!}{7!\times1!\times1!\times1!}$ ways for each permutation of $TFTT$, the total ways in question is
$$
\frac{4!}{3!\times1!} \times \frac{10!}{7!\times1!\times1!\times1!} = {10\choose3} \times P(4,3)
$$
