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Math Help - Counting, Inclusion-Exclusion Rule

  1. #1
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    Counting, Inclusion-Exclusion Rule

    Hi there,

    So here's a problem I've got for homework...

    Seven women and nine men are on the faculty in the mathematics department at a school.
    a) How many ways are there to select a committee of 5 members of the department if at least one woman must be on the committee?
    b) How many ways are there to select a committee of 5 members if at least one woman and one man must be on the committee?
    So for part A I have..
    --> 7 ways to fill first slot with a woman we will call w.
    --> nCr(16,4) ways to fill the next 4 slots
    -----> Minus the combinations containing w = nCr(15,3)
    So the calculation is: 7*(nCr(16,4) - nCr(15,3)) = 9555 (final answer)

    And for part B...
    --> 7 ways to fill first slot with a woman we'll call w
    --> 9 ways to fill second slot with a man we'll call m
    --> nCr(16,3) ways to fill the next 3 slots
    -----> Minus combinations containing w = nCr(15,2)
    -----> Minus combinations containing m = nCr(15,2)
    -----> Plus combinations containing w and m = nCr(14,1)
    So the calculation is: 7*9*(nCr(16,3)-2*nCr(15,2)+nCr(14,1)) = 22932 (final answer)

    These answers just don't seem reasonable... especially since it looks like the additional restrictions that are applied in part B actually result in MORE possible ways to choose the committee instead of less...

    I also tried considering that, in part B, instead of choosing 3 people from 16 to fill the remaining slots, you simply subtract 2 from the number of people (because you exclude w and m from the set of people available to fill the slots), thus the calculation becomes simply 7*9*nCr(14,3), but this yields the exact same answer of 22932. The same applies to part A when I tried applying this logic.

    Where am I going wrong?

    Thanks
    Last edited by Fox2013; April 11th 2011 at 07:28 PM.
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  2. #2
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    Hello, Fox2013!

    Seven women and nine men are on the faculty.
    How many ways are there to select a committee of five people

    (a) if at least one woman must be on the committee?

    \displaystyle\text{With no restrictions, there are: }\:{16\choose5} \:=\:\frac{16!}{5!\,11!} \:=\:4368\text{ possible committees.}

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

    \text{There are: }\:{9\choose5} \:=\:126\text{ all-male committees.}

    Therefore, there are: . 4368 - 126 \:=\:4242\text{ committees with at least one woman.}




    (b) if at least one woman and one man must be on the committee?

    We must not have all-male committees nor all-female committees.

    In part (a), we found that are 126 all-male committees.

    \displaystyle\text{We find that there are: }\:{7\choose5} \,=\,21\text{ all-female committees.}

    Therefore, there are: . 4368 - 126 - 21 \:=\:4221\text{ committees}
    . . \text{with at least one woman and at least one man.}

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