We are given set $S = \{A \in \mathcal P(\{1, \dots 10 \}) \mid 2 \le |A| \le 7 \}$ and have to find the number of maximal and minimal elements in this relation: $(A, \subseteq)$
My idea to solve this is to simply say that all of the two-element sets will be minimal and all seven-element sets will be maximal. And so, the number of minimal elements will be $\frac{10 \cdot 9}{2}$ and the number of maximal elements will be $\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = \frac{10\cdot9\cdot8}{3 \cdot 2}$ Am I missing something or it's that simple?
To make this more precise, note that if there is any $A \in S$ with $|A| > 2$, it cannot be minimal since you can pick up 2-subsets of $A$ that will be "smaller" in the sense of the relation, and similarly if $A| < 7$ it cannot be maximal since there will be $7$-subsets which contain $A$.
In reverse, any 2-subset is minimal since there aren't any more elementary subsets, and any 7-subset is maximal by the reverse logic.
Therefore you are correct. The only thing I would alter is the notation for computing the number of such sets -- I would use $$ \binom{10}{2} = 45 \quad \text{ and } \quad \binom{10}{7}=120, $$ which makes combinatorially clear what you are doing exactly, but they evaluate to the same thing.