Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
P = 1 − (12/12 * 10/11 * 8/10 * 6/9)
In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?
P = 1 - (3/9 * 3/8 * 3/7 * 6/6 * 4!/2!)
Can someone help me spot the difference in probability technique on why 4!/2! multiplier is needed in the second question but 4! isn’t required in the first question?