- #1

- 1,629

- 1

Hello everyone,

I'm trying to understand this example in the book:

It says:

Suppose 2 members of the group of 12 insist on working as a pair--any team must cointain either both, or neither. How many five-person teams can be formed?

What i don't understand is,

For the first part, 10 choose 3, if 2 of the team members must work together, why is it 10 choose 3? Is it becuase u assume you already have chosen 2 of the people, so now you only have to worry about choosing 3 more people to toss in that group?

And for part 2:

its 10 choose 5, is this because you havn't picked anyone yet to be in the group and no restrictions are set so you still have any 5 to choose in the group, so its 10 choose 5?

but if there are 12 people in the group, why isn't it 12 choose 5 and 12 choose 4? how did they get 10?

Because another example in the book, says the total number of teams of 5 is 12 choose 5. Not 10 chose 5

Thanks!

I'm trying to understand this example in the book:

It says:

Suppose 2 members of the group of 12 insist on working as a pair--any team must cointain either both, or neither. How many five-person teams can be formed?

**Solution: Call the 2 memebers of the group that insist on working as a pair A and B. Then any team formed must contain both A and B or neither A nor B. The set of all possible teams can be partitioned into 2 subsets as shown below.**

Because a team that contians both A aand B contains exactly 3 other people from the remaining 10 is the group, there are as many such teams as there are subsets of three people that can be chosen from the remaning ten.

So the answer is 10 choose 3 = 120.

Because a team that contians neither A nor B contains exactly 5 people from the reminaing ten, there are as many such teams as there are subsets of 5 people that can be chosen from the remaning ten.

10 choose 5 = 252.

Because the set of teams that contain both A and B is disjoint from the set of teams that ccontain neither A nor B, by the addition rule:

# of teams containing both A and B or neither A nor B =

# of teams containing both A and B +

# of teams containing neither A nor B.

= 120 + 252 = 372.

Because a team that contians both A aand B contains exactly 3 other people from the remaining 10 is the group, there are as many such teams as there are subsets of three people that can be chosen from the remaning ten.

So the answer is 10 choose 3 = 120.

Because a team that contians neither A nor B contains exactly 5 people from the reminaing ten, there are as many such teams as there are subsets of 5 people that can be chosen from the remaning ten.

10 choose 5 = 252.

Because the set of teams that contain both A and B is disjoint from the set of teams that ccontain neither A nor B, by the addition rule:

# of teams containing both A and B or neither A nor B =

# of teams containing both A and B +

# of teams containing neither A nor B.

= 120 + 252 = 372.

What i don't understand is,

For the first part, 10 choose 3, if 2 of the team members must work together, why is it 10 choose 3? Is it becuase u assume you already have chosen 2 of the people, so now you only have to worry about choosing 3 more people to toss in that group?

And for part 2:

its 10 choose 5, is this because you havn't picked anyone yet to be in the group and no restrictions are set so you still have any 5 to choose in the group, so its 10 choose 5?

but if there are 12 people in the group, why isn't it 12 choose 5 and 12 choose 4? how did they get 10?

Because another example in the book, says the total number of teams of 5 is 12 choose 5. Not 10 chose 5

Thanks!

Last edited: