Combinatorics - Choosing group memebers

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Homework Statement



There is a group of 7 people. How many groups of 3 people can be made from the 7 when 2 of the people refuse to be in the same group?

Homework Equations


nCr


The Attempt at a Solution



Here is what I know:

7C3 gives the total number of groups that can be made.

5C1*2C2 gives the number of groups with the two feuding people; because, 2C2 gives the number of ways to choose both of the feuding people, and 5C1 gives the number of ways to choose any 1 of the none feuding people. Their product gives the number of sets with the two feuding people and 1 of the group of non-feuding people.

Based on this 7C3-5C1*2C2 gives the correct answer.

I understand this is the easiest method to solve the problem; however, I don't understand why 5C2*2C1 doesn't give the number of groups that do not contain the feuding people.

Can anyone explain what I am missing, conceptually?

I understand that (7C3-5C1*2C2)≠(5C2*2C1). But 5C2 gives the number of ways to choose 2 people from the set of non-feuding people. 2C1 gives the number of ways to choose only 1 of the set of feuding people, right/wrong? I thought the product of these two combinations would give the number of sets that do not contain the feuding people, but I am wrong. 5C2*2C1 gives fewer than the actual number of sets that do not contain the feuding people. I gotta be missing something conceptually, what is it?
 

Answers and Replies

  • #2
Dick
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Homework Statement



There is a group of 7 people. How many groups of 3 people can be made from the 7 when 2 of the people refuse to be in the same group?

Homework Equations


nCr


The Attempt at a Solution



Here is what I know:

7C3 gives the total number of groups that can be made.

5C1*2C2 gives the number of groups with the two feuding people; because, 2C2 gives the number of ways to choose both of the feuding people, and 5C1 gives the number of ways to choose any 1 of the none feuding people. Their product gives the number of sets with the two feuding people and 1 of the group of non-feuding people.

Based on this 7C3-5C1*2C2 gives the correct answer.

I understand this is the easiest method to solve the problem; however, I don't understand why 5C2*2C1 doesn't give the number of groups that do not contain the feuding people.

Can anyone explain what I am missing, conceptually?

I understand that (7C3-5C1*2C2)≠(5C2*2C1). But 5C2 gives the number of ways to choose 2 people from the set of non-feuding people. 2C1 gives the number of ways to choose only 1 of the set of feuding people, right/wrong? I thought the product of these two combinations would give the number of sets that do not contain the feuding people, but I am wrong. 5C2*2C1 gives fewer than the actual number of sets that do not contain the feuding people. I gotta be missing something conceptually, what is it?

5C2*2C1 gives you the number of groups that contain exactly one feuding person. It doesn't count the groups that don't have either one.
 
  • #3
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Awesome! I forgot all about the sets with neither. So:

5C2*2C1+5C3 gives the answer I was looking for. Since 5C3 is the number of sets which contain neither person and 5C2*2C1 are the sets with either.

So I needed to think about which groups contained either, and which contained neither.

Thank you!!!!!
 

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