For the first person you have 12 choices, for the second person you have 11 choices, and for the third person you have 10 choices. So the total number of choices is 12 * 11 *10 = 1,320 committees

The first person can be any one of twelve, the second, any one of eleven and the third any one of the remaining 10 so you would (incorrectly) think that you would just multiply those three numbers and get 1,320. However, that does not take into account that the same three people can be picked in more than one sequence. In order to avoid double counting, we must divide that product (1,320) by 1 x 2 x 3, or 6. (The first person can only be picked in one way. The first two can be picked in either of two ways [either being first]. The third gets into more permutations). Therefore, the correct number of different committees of three from a group of 12 is 1,320/6 = 220