In how many ways can a committee of $3$ persons be chosen from a group of $10$ people?

Order does not matter when forming a committee, so this is a combination: $^{10} C_3$. $^{10} C_3 = \dfrac{10!}{3! \cdot 7!} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = \dfrac{720}{6} = 120$. Option C (720) is the trap: that's $^{10} P_3 = 10 \times 9 \times 8$, which counts *ordered* selections (e.g. president, vice-president, secretary as distinct roles). For an unordered committee, divide by $3! = 6$.