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Math Help - Matching problem

  1. #1
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    Matching problem

    My sister had this homework problem, and I'm not quite sure how to solve it.

    There are 6 books that I want to buy, and I can only get 4 of them. How many different combination of books are there for me to choose from?

    Is there a formula for this type of problem?

    Thanks.

    KK
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  2. #2
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    Hello, KK!

    I hope your sister is familiar with factorials . . .

    There are 6 books that I want to buy, and I can only get 4 of them.
    How many different combination of books are there for me to choose from?
    Yes, there is a formula . . .

    This is a Combinations problem -- the order of the books is not important.

    Choosing 4 items from 6, there are: C(6,4) \:=\:\frac{6!}{4!2!}\:=\:15 differemt choices.
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  3. #3
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    That is a "factorial" symbol.

    Examples: . \begin{Bmatrix}3!\:=\:1 \times 2 \times 3 \:=\:6 \\ 4!\:=\:1 \times 2 \times 3 \times 4\:=\:24 \\5! \:= \:1 \times 2 \times 3 \times 4\times 5 = 120 \\ \vdots \\ n!\:=\:1 \times 2 \times 3 \times \cdots \times n\end{Bmatrix}

    By defintion: . 1! = 1,\;\;0! = 1


    So: . \frac{6!}{4!2!}\:=\:\frac{\not{1}\cdot\not{2}\cdot  \not{3}\cdot\not{4}\cdot5\cdot\not{6}^3}{(\not{1} \cdot \not{2}\cdot\not{3}\cdot\not{4})(1\cdot\not{2})} \;= \;15
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