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Math Help - Permutations/Combinations question

  1. #1
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    Permutations/Combinations question

    Given a group of 4 computer science professors and 10 of their TAs:

    (a) how many different 6 person soccer teams can be made?

    (b) how many different 5 person teams can be made that contain at least 1 professor?

    (c) how many different 8 person teams can be made that contain at least 2 professors and at least 4 TAs?

    (d) how many different 4 person teams can be made that contain exactly 2 TAs?

    Please show your work.
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  2. #2
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    Quote Originally Posted by Runty View Post
    Given a group of 4 computer science professors and 10 of their TAs:
    (a) how many different 6 person soccer teams can be made?
    (b) how many different 5 person teams can be made that contain at least 1 professor?
    (c) how many different 8 person teams can be made that contain at least 2 professors and at least 4 TAs?
    (d) how many different 4 person teams can be made that contain exactly 2 TAs?
    Please show your work.
    By all means, do show your work.
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  3. #3
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    Okay, here are the answers I've gotten so far, but I want to make sure if I'm doing them right.

    a) C(14,6)=\frac{14!}{6!8!}=3003

    b) C(14,5)-C(10,5)=\frac{14!}{5!9!}+\frac{10!}{5!5!}=2002-252=1750

    c) C(14,8)-4[C(10,7)]-C(10,8)=\frac{14!}{8!6!}-4\frac{10!}{7!3!}-\frac{10!}{8!2!}=3003-4(120)-45=2478

    d) \frac{10\times 9}{2}\times \frac{4\times 3}{2}=45\times 6=270

    The ones I'm not sure about are the third and fourth ones. Can you confirm if these are correct?
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  4. #4
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    Quote Originally Posted by Runty View Post
    Okay, here are the answers I've gotten so far, but I want to make sure if I'm doing them right.

    a) C(14,6)=\frac{14!}{6!8!}=3003

    b) C(14,5)-C(10,5)=\frac{14!}{5!9!}+\frac{10!}{5!5!}=2002-252=1750

    c) C(14,8)-4[C(10,7)]-C(10,8)=\frac{14!}{8!6!}-4\frac{10!}{7!3!}-\frac{10!}{8!2!}=3003-4(120)-45=2478

    d) \frac{10\times 9}{2}\times \frac{4\times 3}{2}=45\times 6=270
    You are about 1,2,& 4.
    For #3 I have \binom{10}{4}\binom{4}{4} +\binom{10}{5}\binom{4}{3} +\binom{10}{6}\binom{4}{2}
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