dominion wrote:
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?
So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other.
There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - \(C^6_4*(C^2_1)^4\)
look at other problem:
Quote:
Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team?
we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.
I hope it is clearer now.
Wow, makes very clear sense now. Much obliged.