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Math Help - Probability/Combinations - Question

  1. #1
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    Probability/Combinations - Question

    A club consists of 25 married couples. Determine how many ways a committee of 7 people can be formed from these couples if..


    c) at least one person must be a man

    no answer for this question. Could someone please help?


    d) there must be at least 3 women and at least 2 men

    my answer: (25 3) (25 7-2)
    = 25 Cr 3 x 25 Cr 7-2
    = 1,105,598,500

    Is this correct?
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  2. #2
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    Hello, cnmath16!

    A club consists of 25 married couples.
    Determine how many ways a committee of 7 people can be formed from these couples if:

    c) at least one person must be a man.
    There are: . {50\choose 7} \:=\:99,\!884,\!400 possible committees.

    The opposite of "at least one man" is "no men" (all women).

    There are: . {25\choose7} \:=\:480,\!700 all-women committees.
    . . Hence, the rest must contain at least one man.


    Answer: . 99,\!884,\!400 - 480,\!700 \;=\;99,\!403,\!700 committees with at least one man.




    d) there must be at least 3 women and at least 2 men.
    There are three cases: . \begin{array}{ccc}(1) & \text{3 women, 4 men} \\ (2) & \text{4 women, 3 men} \\ (3) & \text{5 women, 2 men} \end{array}

    . . \begin{array}{ccccc}\text{(1) 3 women, 4 men:} & {25\choose3}{25\choose4} &=& 29,\!095,\!000 \\ \\[-3mm]<br />
\text{(2) 4 women, 3 men:} & {25\choose4}{25\choose3} &=& 29,\!095,\!000 \\ \\[-3mm]<br />
\text{(3) 5 women, 2 men:} & {25\choose5}{25\choose2} &=& 15,\!939,\!000  \end{array}


    Answer: . 29,\!095,\!000 + 29,\!095,\!000 + 15,\!939,\!000 \;=\; 74,\!129,\!000 committees.

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