mathmath333
  • mathmath333
Counting question
Mathematics
jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{The number of ways of dividing 10 different balloons }\hspace{.33em}\\~\\ & \normalsize \text{in two groups each containing 5 balloons is ? }\hspace{.33em}\\~\\ \end{align}}\)
Michele_Laino
  • Michele_Laino
I think that the requested number is: \[\Large 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = \left( {\begin{array}{*{20}{c}} {10} \\ 5 \end{array}} \right) \cdot 5!\]
mathmath333
  • mathmath333
but answer given is \(\large \dfrac{10!}{5!5!2!}\)

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Michele_Laino
  • Michele_Laino
I'm sorry!
mathmath333
  • mathmath333
sometimes their is typo in book too
Michele_Laino
  • Michele_Laino
I don't think that it is a typo, I think that my answer is incorrect!
Michele_Laino
  • Michele_Laino
if we have n elements, then for each subset of k elements I have a subset of n-k elements, we have: \[\Large \left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ {n - k} \end{array}} \right)\]
mathmath333
  • mathmath333
this applies for n different elements right
Michele_Laino
  • Michele_Laino
yes! In our case we have n=10 and k=5
mathmath333
  • mathmath333
ohk thanks
Michele_Laino
  • Michele_Laino
I think that the requested number of ways is given by the total number of subset divided by 2=2!
Michele_Laino
  • Michele_Laino
sinceeach way, gives two subsets of 5 elements
Michele_Laino
  • Michele_Laino
since each*
mathmath333
  • mathmath333
ok

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