12 People such as Party A, Party B and Party C draw lots to live in three rooms of abc, with four people in each room, so as to find the probability of Party A, Party B and Party C living in three roo

12 People such as Party A, Party B and Party C draw lots to live in three rooms of abc, with four people in each room, so as to find the probability of Party A, Party B and Party C living in three rooms. This question, first of all, does not require who must live in the a.b.c room, so it is simply regarded as three rooms, that is to say, this question asks the probability that 12 people are divided into three groups, and three people are not in one group, so each group begins to take people. When the first group takes the first person, the probability of choosing any one of them is 3/ 12, and then the second one. So the probabilities are 9/ 1 1, 8/ 10 and 7/9 respectively. Because the special individuals of A, B and C don't have to be in the first person position, two, three and four can be used, so the original probability is better than four, so the probability of the first group is 28/55, and the probability that the first person of the second group is a special individual of A, B and C becomes 2/8. Special individuals can also appear in the number 234, so the probability is still multiplied by 4 to determine the first two groups, and naturally determine the third group, so the probability of the third group is 1. Because these three groups need to meet the conditions at the same time, the probability is multiplied. The answer is 28/55 of the first group times 4/7 of the second group times 1 of the third group, and the final result is 16/0.