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Math Help - [SOLVED] Help with tough Combinations Question

  1. #1
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    [SOLVED] Help with tough Combinations Question

    Hi, I can't seem to answer this question. The answer I get differs from the back of the book and I don't understand why.

    Q: The camera club has 5 members, and the math club has 8. There is only 1 member common to both clubs. In how many ways could a committee of 4 people be formed with a least one member from each club?

    I think the part where I'm going wrong is where the 1 member overlaps. I'm not sure how I'm suppose to deal with it. The answer I get is 470 while the answer at the back is 459...

    If someone could please show me how I'm suppose to go about solving this problem, it would be great!

    THANKS
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  2. #2
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    One way to look at it is to account for that pesky person who is in both clubs.

    \binom{4}{1}\binom{8}{3}+\binom{4}{2}\binom{8}{2}+  \binom{4}{3}\binom{8}{1}+\binom{7}{4}\binom{5}{0}

    Draw a Venn and it will be easier to see.
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  3. #3
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    Quote Originally Posted by galactus View Post
    One way to look at it is to account for that pesky person who is in both clubs.

    \binom{4}{1}\binom{8}{3}+\binom{4}{2}\binom{8}{2}+  \binom{4}{3}\binom{8}{1}+\binom{7}{4}\binom{5}{0}

    Draw a Venn and it will be easier to see.
    I'm not sure I'm understanding this correctly. So for part of the question, you include that pesky person on one side and then on the other for the last part of the question. You get the correct answer.

    But what if I decided to include the pesky person on the other side first then the first side. As in:

    \binom{7}{1}\binom{5}{3}+\binom{7}{2}\binom{5}{2}+  \binom{7}{3}\binom{5}{1}+\binom{4}{4}\binom{8}{0}<br />

    The answer then becomes 456
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  4. #4
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    Quote Originally Posted by MATHDUDE2 View Post
    Q: The camera club has 5 members, and the math club has 8. There is only 1 member common to both clubs. In how many ways could a committee of 4 people be formed with a least one member from each club?
    answer at the back is 459...
    There are really 12 people total.
    So there are a total of \binom{12}{4} ways to select the committee.
    There \binom{7}{4}+\binom{4}{4} ways the committee will not have someone from each club.

    \binom{12}{4}-\left(\binom{7}{4}+\binom{4}{4}\right)=459
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  5. #5
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    Quote Originally Posted by Plato View Post
    There are really 12 people total.
    So there are a total of \binom{12}{4} ways to select the committee.
    There \binom{7}{4}+\binom{4}{4} ways the committee will not have someone from each club.

    \binom{12}{4}-\left(\binom{7}{4}+\binom{4}{4}\right)=459
    WOW I can't believe I didn't see the indirect method as a viable option. Thanks! Still curious as to how to solve it using the direct method but at least I know how to solve the question now. Thanks everyone!
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  6. #6
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    Here's how I solved it directly. There are 4 people in club 1 and 7 in club 2 and the one guy on the side who can count for both. Now, we either choose to include the one guy in the committee or not. If we don't choose him, then there are

    \sum_{k=1}^3 {4 \choose k} {7 \choose {4-k}}

    ways to pick the committee. That is, we would like to pick out k from one group and the rest from the other, and note that k is from 1 to 3, so one person has to be chosen from each group.

    Now we concern ourselves with the person who's in both camps. We've picked him, so we need to pick 3 other people. They can belong to any club since the one person counts from both camps. Thus we add

    {11 \choose 3}

    to the above. Evaluating that will give you 459 as the book says.
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