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Math Help - Combinatorics

  1. #1
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    Combinatorics

    A professor of a class with 20 students must select 4 people to enter a math fair and 3 people to enter a contest on "Math in the News." In how many ways may the professor make the selections if:
    1.) the two groups may not overlap
    2.) the may overlap?
    3.) they must overlap?
    4.) exactly two students enter both the math fair and the contest?
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  2. #2
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    Quote Originally Posted by fifthrapiers View Post
    A professor of a class with 20 students must select 4 people to enter a math fair and 3 people to enter a contest on "Math in the News." In how many ways may the professor make the selections if:
    1.) the two groups may not overlap
    2.) the may overlap?
    3.) they must overlap?
    4.) exactly two students enter both the math fair and the contest?
    In what follows, let C(n,k) denote the combination of n taking k.

    1) C(20,4)*C(16,3), choose the math fair then the Math News from who is left.

    2) C(20,4)*C(20,3), choose the math fair then the Math News from the same pool.

    3) Is the difference: #2 - #1

    4) C(20,2)*C(18,2)*C(16,1)= [20!]/([2!]^2[15!]).

    Here another way to think about #4. Suppose we have 2 B's, 2 F's, 1 M & 15 X's. How many ways can we form a string of the letters? If a person in our list get a B, that person is in both; F means math fair; M means Math News; and X means not chosen.

    CORRECTED! Thanks Soroban
    Last edited by Plato; September 12th 2006 at 01:16 PM.
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  3. #3
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    Hello, Plato!

    I agree with your answers . . . except #4.
    (And I bet it was just a teensy oversight)


    4) C(20,2)ĚC(18,2)ĚC(16,2)

    There are: C(20,2) ways to choose the two who are in both activities.

    Then there are: C(18,2) ways to choose two more to enter the fair from the other 18 students

    and: C(16,1) ways to choose one more to enter the contest from the other 16 students.

    Answer: .C(20,2)ĚC(18,1)ĚC(16,1) .= .465,120

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