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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{The number of ways of dividing 10 different balloons }\hspace{.33em}\\~\\ & \normalsize \text{in two groups each containing 5 balloons is ? }\hspace{.33em}\\~\\ \end{align}}\)

  2. Michele_Laino
    • one year ago
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    I think that the requested number is: \[\Large 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = \left( {\begin{array}{*{20}{c}} {10} \\ 5 \end{array}} \right) \cdot 5!\]

  3. mathmath333
    • one year ago
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    but answer given is \(\large \dfrac{10!}{5!5!2!}\)

  4. Michele_Laino
    • one year ago
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    I'm sorry!

  5. mathmath333
    • one year ago
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    sometimes their is typo in book too

  6. Michele_Laino
    • one year ago
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    I don't think that it is a typo, I think that my answer is incorrect!

  7. Michele_Laino
    • one year ago
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    if we have n elements, then for each subset of k elements I have a subset of n-k elements, we have: \[\Large \left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ {n - k} \end{array}} \right)\]

  8. mathmath333
    • one year ago
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    this applies for n different elements right

  9. Michele_Laino
    • one year ago
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    yes! In our case we have n=10 and k=5

  10. mathmath333
    • one year ago
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    ohk thanks

  11. Michele_Laino
    • one year ago
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    I think that the requested number of ways is given by the total number of subset divided by 2=2!

  12. Michele_Laino
    • one year ago
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    sinceeach way, gives two subsets of 5 elements

  13. Michele_Laino
    • one year ago
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    since each*

  14. mathmath333
    • one year ago
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    ok

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