How many ways are there to form a committee

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The discussion revolves around forming committees from a club of 5 men and 6 women, addressing two specific scenarios. For the first scenario, the correct approach to form a committee of 6 people, considering two women who refuse to serve together, involves using the subtraction principle, leading to a total of 336 combinations. In the second scenario, forming a committee of 4 men and 3 women, the calculation yields 40 valid combinations after accounting for the two men who cannot be together. Additionally, a related probability problem about arranging flowers highlights the challenge of applying mathematical formulas versus visualizing the arrangement. Overall, the participants seek clarity on using combinatorial principles effectively to solve these problems.
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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!
 
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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?
 
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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 alot.
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.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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