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Math Help - Basic Counting Principles

  1. #1
    DBA
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    Basic Counting Principles

    Given nine different English books, seven different French books, and five different German books.

    How many ways are there to make a row of three books in which exactly one language is missing (the order of the three books makes a difference)?

    The answer is
    3*[9*8* (7+5) + 7*6*(9+5) + 5*4*(9+7)]

    I do not understand how to get there.

    I have 3 locations in the row _ _ _

    In the case I use only French and German books I have
    7+5=12 books availabe to place
    I do not kow where the 9*8 come from in the expression 3*(9*8*(7+5))

    I seem to have 3 subsets which are added (French/German, Engl/German, Engl/French).

    Can someone help me out? Thanks
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  2. #2
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    Quote Originally Posted by DBA View Post
    Given nine different English books, seven different French books, and five different German books.
    How many ways are there to make a row of three books in which exactly one language is missing (the order of the three books makes a difference)?
    The answer is 3*[9*8* (7+5) + 7*6*(9+5) + 5*4*(9+7)]
    It is  3\cdot[9\cdot8\cdot (7+5) + 7\cdot6\cdot(9+5) + 5\cdot4\cdot(9+7)]

    Consider the case in which we use two English texts. That can happen in 9\cdot 8 ways. Then select either one French text or one German text to go with the two English books.
    That can be done in 9\cdot 8\cdot(7+5) ways. Then there are 3 places to put the single book.
    That gives 3\cdot[9\cdot 8\cdot(7+5)] to form a line of two English texts, one French text, or a line of two English texts, one German text.
    Now you have two more cases.
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  3. #3
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    Some hint -
    "In the case I use only French and German books I have
    7+5=12 books availabe to place"

    But there is an added condition - there has to be atleast 1 book of French and German each.

    Can you work fwd now?
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  4. #4
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    Hello, DBA!

    Given: 9 distinct English books, 7 distinct French books, and 5 distinct German books.

    How many ways are there to make a row of three books
    in which exactly one language is missing?
    (The order of the three books makes a difference.)

    \text{The answer is: }\;3\cdot\bigg[9\cdot8\cdot (7+5) + 7\cdot6\cdot(9+5) + 5\cdot4\cdot(9+7)\bigg]

    Their answer is correct . . . but why did they write it that way??


    Suppose we want two English books and one Other.
    . . There are 9 English and 12 Others.
    . . There are: . {9\choose2}{12\choose1} choices.
    . . The three books can be arranged on the shelf in 3! ways.

    There are: . 3!{9\choose2}{12\choose1} ways to have two English books and one Other.


    Suppose we want two French books and one Other.
    . . There are 7 French and 14 Others.
    . . There are: . {7\choose2}{14\choose1} choices.
    . . The three books can be arranged on the shelf in 3! ways.

    There are: . 3!{7\choose2}{14\choose1} ways to have two French books and one Other.


    Suppose we want two German books and one Other.
    . . There are 5 German and 16 Others.
    . . There are: . {5\choose2}{16\choose1} choices.
    . . The three books can be arranged on the shelf in 3! ways.

    There are: . 3!{5\choose2}{16\choose1} ways to have two German books and one Other.


    Hence, there are: . \displaystyle 3!{9\choose2}{12\choose1} + 3!{7\choose2}{14\choose1} + 3!{5\choose2}{16\choose1}

    . . . . . . . . . . . \displaystyle =\;3!\bigg[{9\choose2}\!\cdot\112 + {7\choose2}\!\cdot\!14 + {5\choose2}\!\cdot\116\bigg]\text{ ways} .[1]


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    At this point, I would have cranked out the answer:

    . . 6\cdot\bigg[\frac{9\cdot8}{2}\cdot12 + \frac{7\cdot6}{2}\cdot14 + \frac{5\cdot4}{2}\cdot16\bigg] \;=\;6(886) \;=\;5316


    Or left it as [1] so others might see how I got the answer.


    Their "3" out front can be explained,
    . . but I think it is confusing and misleading.
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