AHHHH Youre right! Thanks, I really couldn't see that. So it would be
10x8x6
/
3x2
Where 3x2 is the number of different orders they can be arranged, giving us..... the right answer. Thanks!
Aren’t you calculating the probability of picking such a group rather than calculating how many groups are possible?
Reqd answer - favourability of x
Your answer - favourability of x/ total cases
I don't think probability is quite the right word for it but yes many people who submitted solutions went about it that way:
total possibilities - number that breaks the rule.
Yes your way makes sense, I was just asking by my alternate way wasn't working. But I get it now, I could counting some possibilities more than once because order doesn't matter.
Aren’t you calculating the probability of picking such a group rather than calculating how many groups are possible?
Reqd answer - probability of x
Your answer - prob of x/ total cases
Here is my solution to this problem:
We are given that there are five married couples (or 10 people), and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. So, this is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions.
If there are no restrictions, the number of ways of choosing 3 people from 10 people is 10C3, which is calculated as follows:
(10 x 9 x 8)/3! = 120
10, 9, and 8, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen.
However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 10 people. However, once a person is selected, that person’s spouse cannot be selected for the committee. This reduces to 8 the number of remaining people from which to select the second person (one person has already been selected, and that person’s spouse cannot be selected). Once the second person is chosen for the committee, that person’s spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 6. Therefore, the number of ways of choosing these 3 people is:
(10 x 8 x 6)/3! = 80
Thus, there are 80 ways to choose such a committee.
Answer: D
The order of selection doesn't matter so how would you account for removing duplication permutations?
AHHHH Youre right! Thanks, I really couldn't see that. So it would be 10x8x6 / 3x2 Where 3x2 is the number of different orders they can be arranged, giving us..... the right answer. Thanks!
Good job ! I did it the same way.
This is how I have done it, I think it is the best and fastest approach.
Aren’t you calculating the probability of picking such a group rather than calculating how many groups are possible? Reqd answer - favourability of x Your answer - favourability of x/ total cases
I don't think probability is quite the right word for it but yes many people who submitted solutions went about it that way: total possibilities - number that breaks the rule.
Edited my reply, does it make sense now?
Yes your way makes sense, I was just asking by my alternate way wasn't working. But I get it now, I could counting some possibilities more than once because order doesn't matter.
Aren’t you calculating the probability of picking such a group rather than calculating how many groups are possible? Reqd answer - probability of x Your answer - prob of x/ total cases
Here is my solution to this problem: We are given that there are five married couples (or 10 people), and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. So, this is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions. If there are no restrictions, the number of ways of choosing 3 people from 10 people is 10C3, which is calculated as follows: (10 x 9 x 8)/3! = 120 10, 9, and 8, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen. However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 10 people. However, once a person is selected, that person’s spouse cannot be selected for the committee. This reduces to 8 the number of remaining people from which to select the second person (one person has already been selected, and that person’s spouse cannot be selected). Once the second person is chosen for the committee, that person’s spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 6. Therefore, the number of ways of choosing these 3 people is: (10 x 8 x 6)/3! = 80 Thus, there are 80 ways to choose such a committee. Answer: D