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Math Help - !!Need someone to check answers - Combinatorics Problems

  1. #1
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    !!Need someone to check answers - Combinatorics Problems

    36. Solve for n.

    a) P (n,5) = 20P (n,3)

    No clue how to do this; - please help me solve!

    37. A library shelf contains 7 different books. In how many ways can they be arranged if

    a) there are no restrictions

    Answer: They can be arranged in 5040 different ways 7! = 5040

    b) the Mathematics of Data Management textbook must be first

    Answer: 6! x 1 = 720 different ways

    c) the 3 mathematics textbooks must be together

    Answer: - when the 3 textbooks go together, they can be counted as one textbook
    Therefore, 4! x 1 = 24 different ways.
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  2. #2
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    Quote Originally Posted by cnmath16 View Post
    36. Solve for n.

    a) P (n,5) = 20P (n,3)

    No clue how to do this; - please help me solve!
    What does P(n,5) mean?

    Quote Originally Posted by cnmath16 View Post
    37. A library shelf contains 7 different books. In how many ways can they be arranged if

    a) there are no restrictions

    Answer: They can be arranged in 5040 different ways 7! = 5040

    b) the Mathematics of Data Management textbook must be first

    Answer: 6! x 1 = 720 different ways
    a) is correct

    b) is correct


    Quote Originally Posted by cnmath16 View Post
    c) the 3 mathematics textbooks must be together

    Answer: - when the 3 textbooks go together, they can be counted as one textbook
    Therefore, 4! x 1 = 24 different ways.

    Well, and 4! means, that you have this event:

    Mathbooks, Book A, Book B, Book C, Book D

    Mathbooks, Book A, Book C, Book B, Book D

    ...

    What about
    Book A, Book A, Book C, Book B, Book D?

    There are

    Mathbooks, A,B,C,D
    A,Mathbooks,B,C,D
    A, B, mathbooks, C, D
    A,B, C, mathbooks, D
    A,B,C,D, mathbooks

    5 different ways, thats why I think the soluition is 4!*5
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  3. #3
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    37. c) Answer: 5! x 3!

    Do you see why?
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  4. #4
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    Quote Originally Posted by awkward View Post
    37. c) Answer: 5! x 3!

    Do you see why?
    Yes i do.
    First: i wrote

    Mathbooks, A,B,C,D
    A,Mathbooks,B,C,D
    A, B, mathbooks, C, D
    A,B, C, mathbooks, D
    A,B,C,D, mathbooks

    i forgot, that there are no restriction, means for A,B,C,D there are 4 ! possobilities.

    Adding the mathbooks you got A,B,C,D,math => 5!

    there are three mathbooks
    possible orders are

    math1, math2, math3
    math1, math3, math2
    math2, math3, math 1
    math2, math1, math 3
    math3, math1, math 2
    math3, math2, math 1

    => 3!


    => 5! times 3!

    Sorry for misleading you, cnmath16
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