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Math Help - Combinations problem

  1. #1
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    Combinations problem

    Hello!

    I'm doing the question 1.10 from the Ross text and can't figure out how to get the answer they provide. The question is:

    "A woman has 8 friends, of whom she will invite 5 to a tea party. How many choices has she if 2 of the friends are feuding and will not attend together? How many choices has she if 2 of her friends will only attend together?"

    I know that to select 5 out of the 8, it's 8 choose 5 or (8*7*6*5*4)/(5*4*3*2*1). But I can't figure out how to make sure that friends A and B are not together. I know that it's not necessarily the case that these women will even be part of the 5. Can you suggest the next step?

    Thanks!
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  2. #2
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    Re: Combinations problem

    Wait, I think I might have it. Do you do 8 choose 5, and then 6 choose 4 and 2 choose 1 and subtract the latter from the former (for the first part of the problem)?
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  3. #3
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    Re: Combinations problem

    Quote Originally Posted by Lintu View Post
    "A woman has 8 friends, of whom she will invite 5 to a tea party. How many choices has she if 2 of the friends are feuding and will not attend together? How many choices has she if 2 of her friends will only attend together?"
    Let's do the second on first.
    I assume the questions are not related.
    Invite the two and choose three others.
    That can be done in \binom{6}{3}.

    Now the first part is the opposite. Do you see why?

    BTW: There are several authors with surname Ross.
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  4. #4
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    Re: Combinations problem

    The questions are together in the text A First Course in Probability, Sheldon Ross.

    For the second question, we could choose 5 from the 8 that do not include those 2 people (so 6 choose 5), or we could choose 3 of the 6 and then 2 of the 2 remaining. That would be (6*5*4)/(3*2*1) + (6*5*4*3*2)/(5*4*3*2*1). Right?
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  5. #5
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    Re: Combinations problem

    Quote Originally Posted by Lintu View Post
    The questions are together in the text A First Course in Probability, Sheldon Ross.
    For the second question, we could choose 5 from the 8 that do not include those 2 people (so 6 choose 5), or we could choose 3 of the 6 and then 2 of the 2 remaining. That would be (6*5*4)/(3*2*1) + (6*5*4*3*2)/(5*4*3*2*1). Right?
    For the first the correct answer is \binom{8}{5}-\binom{6}{3}
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  6. #6
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    Re: Combinations problem

    Can you help me get to the logic behind that method? I've gotten to the right number answers, now, but I want to understand the most simple and logical way to get to the solution.

    for 8 choose 5 minus 6 choose 3, it looks to me that you take the selection of 5 from 8 and then take out the selection of 3 from 6, because the remainder would be taking 2 out of the 2 feuding friends.

    And for the other, it's 6 choose 5 + 6 choose 3, because you could either choose 5 out of the 6 non friends, or you could select 3 from the non-friends and 2 of the friends, right?

    (Also, I would have thought this belonged in probability rather than a university-level forum! Sorry about that.)
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  7. #7
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    Re: Combinations problem

    Quote Originally Posted by Lintu View Post
    Can you help me get to the logic behind that method? I've gotten to the right number answers, now, but I want to understand the most simple and logical way to get to the solution.
    There are \binom{6}{3} ways to put those to hateful people on the same list of invitations.

    So there are \binom{8}{5}-\binom{6}{3} to exclude having those hateful people on the same list of invitations.
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