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abz_tech

  • 4 years ago

probability(choosing) question: if (5C0)+(5C1)+(5C2)+(5C3)+(5C4)+(5C3)=2^k, find k.

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  1. abz_tech
    • 4 years ago
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    \[If\left(\begin{matrix}5 \\ 0\end{matrix}\right)+\left(\begin{matrix}5 \\ 1\end{matrix}\right)+\left(\begin{matrix}5 \\ 2\end{matrix}\right)+\left(\begin{matrix}5 \\ 3\end{matrix}\right)+\left(\begin{matrix}5 \\ 4\end{matrix}\right)+\left(\begin{matrix}5 \\ 3\end{matrix}\right) = 2^{k}, find k\]

  2. phi
    • 4 years ago
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    isn't the last term 5C5 ? otherwise you don't get an even number. to solve, simplify the terms on the left to numbers, add them up, and expects a power of 2. For example 5C0= 1. continue for 5C1, etc

  3. abz_tech
    • 4 years ago
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    nope, in book is 5C3 (the last one) prolly miss print then i guess =/

  4. cyter
    • 4 years ago
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    \[\sum_{c=0}^{6}{\left(\begin{matrix}5 \\ c\end{matrix}\right)}=2^k\] \[32=2^k\rightarrow 2^5=2^k\] k=5

  5. abz_tech
    • 4 years ago
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    0_o

  6. cyter
    • 4 years ago
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    If last term isn't 5C5 then k=5.39232

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