How many ways are there to form a committee

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Homework Help Overview

The discussion revolves around combinatorial problems involving the formation of committees from a group of men and women, with specific restrictions on membership. The original poster presents two scenarios: forming a committee of 6 people with restrictions on women and forming a committee of 4 men and 3 women with restrictions on men. Additionally, a related probability problem about arranging flowers is introduced.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply combinations to solve the problems but expresses uncertainty about their calculations. They question their understanding of the subtraction principle and seek clarification on how to approach the problems correctly.

Discussion Status

Some participants have offered hints regarding the subtraction principle, which the original poster acknowledges as helpful. There is an ongoing exploration of different methods to solve the problems, including a transition to a new related problem about arranging flowers. The discussion reflects a mix of attempts and guidance without reaching a definitive conclusion.

Contextual Notes

The original poster indicates that the solutions they arrived at do not match the answers provided in the book, suggesting a potential misunderstanding of the problem constraints or the application of combinatorial principles.

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Homework Statement



A certain club consists of 5 men and 6 women.

a) How many ways are there to form a committee of 6 people if a certain pair of women refuse to serve on the same committee?

b) How many ways are there to form a committee of 4 men and 3 women if two of the men refuse to serve on the same committee?

Homework Equations



Permutations & Combinations

P(n,k) = n(n - 1)(n - 2)...(n - k + 1)

C(n,k) = P(n,k)/k!

The Attempt at a Solution



a) Well, a pair = 2, so I intuitively want to exclude one of the women if she will not be in the committee when the other is in the committee. That gives 10 individuals (men and women). C(10,6) = 210

But this is not the answer in the book, unfortunately, so I know I'm not on the right track.

b) C(6,3) = 20 would represent the combinations of the women's committee.

For the men's 4 member committee, I feel like I want to exclude one of the 5 men because, like in the question above, there are two that cannot be in one committee together. That would leave 4 men to fill 4 spaces, so all the possible combinations are...1.

C(6,3) x 1 = 20

But this too is very wrong according to the book. :-\ Can anyone help me to conceptualize this problem correctly? Thanks a lot I really appreciate it!
 
Last edited:
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Do you know what the subtraction principle is? Try using it for both problems.
 
Thank you for your response! It took a while, but with your hint, I did eventually get it. Here is what I did:

a) C(11,6) - C(9,5) = 336

b) C(6,3) x [C(5,4) - C(3,2)] = 40

Thanks again. :-)

BTW, this is another probability problem that, for some reason, I am really hung up on. Can anyone offer any guidance?

Suppose you have 8 red flowers and 8 white flowers, and assume that the flowers are indistinguishable except by their color.
c) How many ways are there to arrange 5 of the red flowers and 5 of the
white flowers in a row if all of the red flowers must be kept together?
 
Last edited:
Can you solve the problem below?

How many ways are there to arrange 1 red flower and 5 white flowers in a row?

-----

Does it matter that you have 5 red flowers in your problem? Could it be any positive number?
 
Thank you again, Mattmns. That really helped a lot.
The answer is definitely 6. But I wasn't able to come up with the answer by figuring it out mathematically (e.g. by applying a formula), I had to draw a picture on a piece of paper. Do you know of any way to figure the problem out purely mathematically without conceptualizing it "manually?" Maybe the permutation theorem would work here somehow, but I'm not really sure how.
 

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