In how many ways can a committee of 3 people be chosen from a group of 8 people?

The order of selection does not matter for a committee, so use a combination: $\binom{8}{3}=\frac{8!}{3!\,5!}=\frac{8\cdot7\cdot6}{3\cdot2\cdot1}=56$, choice G. Choice H ($336$) is the permutation $8\cdot7\cdot6$, which counts each group $3!=6$ times because it treats different orders of the same three people as different committees. Choice F ($24$) is $8\cdot3$, a product that does not model any selection. Choice J ($512$) is $8^3$, which would count ordered choices with repetition allowed, impossible for a committee of distinct people.