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Math Help - More Data .. Can you figure it out?

  1. #1
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    Post More Data .. Can you figure it out?

    A school has 480 girls and 520 boys. how many committees of 5 can u make if at least one boy must be on each committee.
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  2. #2
    Junior Member roy_zhang's Avatar
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    There are two ways to do it:

    1) Simpler Way (solving it indirectly):

    Find the total number of committees and then subtracts the number of committees which are all girls, as
    \binom{1000}{5}-\binom{480}{5}

    OR

    2) Solving it directly:
    \binom{520}{1}\binom{480}{4}+\binom{520}{2}\binom{  480}{3}+\binom{520}{3}\binom{480}{2}+\binom{520}{4  }\binom{480}{1}+\binom{520}{5}
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  3. #3
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    Hello, wikji!

    Who assigned this problem?
    . . In fact, who wrote it?


    A school has 480 girls and 520 boys. How many committees of five
    can be made if at least one boy must be on each committee?

    There are: . {1000\choose5} \:\approx\:82.5\times 10^{11} possible committes.


    The opposite of "at least one boy" is "no boys" (all girls).

    There are: . {480\choose5} \;\approx\:2.1\times10^{11} all-girl committees.


    There are about: . 8\times 10^{12} (eight trillion) committees with at least one boy.

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, wikji!

    Who assigned this problem?
    . . In fact, who wrote it?



    There are: . {1000\choose5} \:\approx\:82.5\times 10^{11} possible committes.


    The opposite of "at least one boy" is "no boys" (all girls).

    There are: . {480\choose5} \;\approx\:2.1\times10^{11} all-girl committees.


    There are about: . 8\times 10^{12} (eight trillion) committees with at least one boy.

    How do u get this ... Sry but im the worst math student of all time ... here are about: .8\times 10^{12} (eight trillion) committees with at least one boy.
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