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Math Help - Combination/ Permutation Question

  1. #1
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    Combination/ Permutation Question

    I have a few questions i don't understand T_T

    1. A committe of 4 people is to be chosen form a group of 8 people- 4 women and 4 men. How many ways can the committee be chosen so as to include exactly 3 women.

    The answer is 70 but i keep getting 96. I don't understand how they got 70 =3=


    2. From a deck of 52 cards, the 12 face cards and 4 aces are removed. From the 16 cards, 4 are chosen. How many different ways are there to choose the numbers?

    3. In the Lotto 535 lottery, you must choose 4 numbers from 1 to 35 and one number from 1 to 25. How many different ways are there to choose the numbers?

    4. Solve for n
    nP2= 12
    The answer is 4 but i get 5 =3=

    n!/ n-2! = 12
    then i expanded and canceled out that n! leaving me (n-2)(n-1) = 12
    i expanded it then i get n^2- 3n+2 = 12
    i got n= 5 and -2
    D:
    i don't know what i did wrong.

    help anyone? (:
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  2. #2
    MHF Contributor matheagle's Avatar
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    For (1) I get 16 since {4\choose 3}{4\choose 1}=(4)(4)=16


    (4) IT's n(n-1)=12 producing n^2-n-12=(n-4)(n+3)=0 so n=4
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  3. #3
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    Quote Originally Posted by matheagle View Post
    I get 16
    Ohh. o_o
    weird T_T
    i don't like these questions
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  4. #4
    MHF Contributor matheagle's Avatar
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    so how did you get 96 and who says its 70?
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  5. #5
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    Quote Originally Posted by matheagle View Post
    so how did you get 96 and who says its 70?
    O; oh em gee. nvm it's 16 D:
    i was looking at another question =3=

    buut. i think using the 8 instead the 4 thats why i was getting such big numbers.
    but how do you know it's a combination or a permutation from

    ( 4 3 ) (4 1)
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  6. #6
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    I thought it was 16 too.

    {8\choose 4}=70, which would be the number of ways to select 4 people out of 8 disregarding gender. Maybe you're reading the wrong answer?

    Haha.. you just beat me to it.

    It's a combination, because the order doesn't matter.
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  7. #7
    MHF Contributor matheagle's Avatar
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    you want to pick 3 women from a group of 4.

    that's {4\choose 3}=4

    you can write out ABCD women, and see there are 4 ways.
    BUT the easier way is to cast off one and see there are 4 cast offs, leaving you with 3 women
    ABC sending D home alone
    ABD sending off C
    ACD sending off B
    BCD sending off A

    same with 1 man from 4, there are 4 men to choose from.
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  8. #8
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    Quote Originally Posted by matheagle View Post
    you want to pick 3 women from a group of 4.

    that's {4\choose 3}=4

    you can write out ABCD women, and see there are 4 ways.
    BUT the easier way is to cast off one and see there are 4 cast offs, leaving you with 3 women
    ABC sending D home alone
    ABD sending off C
    ACD sending off B
    BCD sending off A

    same with 1 man from 4, there are 4 men to choose from.
    okkay. i think i got it xP
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  9. #9
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    2. It looks like their just asking for the number of ways to choose 4 cards out of the remaining 16, although I'm not too sure what is meant by "numbers."

    Anyway, {16 \choose 4} =...

    3. It's the number of ways to choose 4 numbers out of 35 multiplied by the number of ways to choose 1 number out of 25
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  10. #10
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    Quote Originally Posted by downthesun01 View Post
    2. It looks like their just asking for the number of ways to choose 4 cards out of the remaining 16, although I'm not too sure what is meant by "numbers."

    Anyway, {16 \choose 4} =...
    okay...
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