Q.(i) Evaluate $^{10}C_{4}$.

Since committee membership carries no ranking among the 4 chosen members, this is a combination problem, and part (ii) is numerically identical to part (i).

(i) $^{10}C_{4} = \dfrac{10!}{4!\,(10-4)!} = \dfrac{10!}{4!\,6!}$. Cancelling the $6!$ that is common to $10!$ and the denominator: $\dfrac{10\times9\times8\times7}{4\times3\times2\times1} = \dfrac{5040}{24} = 210$.

(ii) A committee has no internal ranking among its 4 members (unless the question separately assigns roles), so selecting 4 members from 10 candidates is a combination, not a permutation. The answer is therefore $^{10}C_{4}=210$, the same value as part (i). …

Using $^{10}P_{4}=5040$ instead of $^{10}C_{4}=210$ for a plain committee-selection question — a permutation would only be correct if the 4 selected members were also being assigned …