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Math Help - Combinatorics

  1. #1
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    Combinatorics

    A committee of 6 is chosen from 10 Mathematics majors and 7 Physics majors so as to contain at least 3 Mathematics majors and at least 2 Physics majors . Show that there are 7800 different ways this can be done if two particular Physics majors refuse to stay on the same committee .


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  2. #2
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    Re: Combinatorics

    Suppose the first person you choose is a math major- there are 6 ways to do that. Suppose the second and third persons are also math majors. There are 5 and 4 ways of doing that so there are 6(5)(4)= \frac{6!}{3!} ways to choose the three math majors. Do the same to find the number of ways to choose 2 physics majors. That's 5 of your commitee. The last two can be chosen from any of the remaining 7+ 5= 12 students. There are 12(11)= \frac{12!}{10!} ways to do that. Multiply them altogether.

    Finally, divide by the number of ways to order "MMMPPEE" ("M" for "math major", "P" for "physics major", "E" for "either") so you are not counting different orders of choosing the same people.
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  3. #3
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    Re: Combinatorics

    Quote Originally Posted by mathlover14 View Post
    A committee of 6 is chosen from 10 Mathematics majors and 7 Physics majors so as to contain at least 3 Mathematics majors and at least 2 Physics majors . Show that there are 7800 different ways this can be done if two particular Physics majors refuse to stay on the same committee .
    Here is another way. The total minus the number containing both who will not serve together.
    \sum\limits_{k = 3}^4 \binom{10}{k} \binom{7}{6-k} }-\sum\limits_{k = 3}^4 \binom{10}{k} \binom{5}{4-k} }
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  4. #4
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    Re: Combinatorics

    Hello, mathlover14!

    A committee of 6 is chosen from 10 Math majors and 7 Physics majors
    so as to contain at least 3 Math majors and at least 2 Physics majors.

    Show that there are 7800 different ways this can be done
    if two particular Physics majors refuse to be on the same committee.

    The committee must contain: . \begin{Bmatrix} \text{3 Math, 3 Physics} & [1] \\ \text{or} \\\text{4 Math, 2 Physics} & [2] \end{Bmatrix}

    \text{[1] 3 Math, 3 Physics}: \;{10\choose3}{7\choose3} \:=\:4200\text{ ways.}

    \text{[2] 4 Math, 2 Physics}:\;{10\choose4}{7\choose2} \:=\:4410\text{ ways.}

    \text{Hence, there are: }\:4200 + 4410 \:=\:\boxed{8610}\text{ possible committees.}



    \text}Suppose }A\text{ and }B\text{ are the two unfriendly Physics majors.}

    \text{In how many ways can they be }together\text{ on the same committee?}


    \text{[1] 3 Math, 3 Physics}

    . . \text{We have: }\;\begin{array}{cc}\text{Math} & \text{Physics} \\ \{\_\:\_\:\_\} & \{A\:B\:\_\} \end{array}

    . . \text{There are: }\:{10\choose3}{5\choose1} \:=\:600\text{ ways.}


    \text{[2] 4 Math, 2 Physics}

    . . \text{We have: }\;\begin{array}{cc}\text{Math} & \text{Physics} \\ \{\_\:\_\:\_\:\_\} & \{A\:B\} \end{array}

    . . \text{There are: }\:{10\choose4}{5\choose0} \:=\:210\text{ ways.}


    \text{Hence, }A\text{ and }B\text{ can be together in: }\:600 + 210 \:=\:\boxed{810}\text{ ways.}


    \text{Therefore, }A\text{ and }B \text{ are }not\text{ together in: }\:8610 - 810 \:=\:\boxed{7800}\text{ ways.}

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