In how many ways can a $3$-person committee be selected from a group of $9$ students? (The order of selection does not matter.)

Since the committee is unordered, this is the number of combinations of $9$ objects taken $3$ at a time: $\binom{9}{3} = \dfrac{9!}{3! \, 6!} = \dfrac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} = \dfrac{504}{6} = 84$. - **C** ($504 = 9 \cdot 8 \cdot 7$) is the count if order **did** matter — i.e. number of *permutations* of 3 from 9. Dividing by $3!$ corrects for the over-counting of orderings within a committee. - **D** ($720 = 6!$) is unrelated. - **A** ($27 = 9 \cdot 3$) attempts a multiplication-principle shortcut that doesn't apply. Memorable check: $\binom{n}{k} = \binom{n}{n-k}$, so $\binom{9}{3} = \binom{9}{6}$ — both equal 84.