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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{How many differenrt signals can be made by hoisting }\hspace{.33em}\\~\\ & \normalsize \text{5 different coloured flags one above the other ,when }\hspace{.33em}\\~\\ & \normalsize \text{any number of them may be hoisted at a time }\hspace{.33em}\\~\\ & a.)\ 2^{5} \hspace{.33em}\\~\\ & b.)\ ^{5}P_{5} \hspace{.33em}\\~\\ & c.)\ 325 \hspace{.33em}\\~\\ & d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    is hoisting 0 flags considered a signal or no?

  3. Michele_Laino
    • one year ago
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    I think that the requested number is given by the subsequent computation: \[\Large \left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 5 \end{array}} \right) = {\left( {1 + 1} \right)^5}\]

  4. anonymous
    • one year ago
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    @Michele_Laino you forgot 5C0

  5. ganeshie8
    • one year ago
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    i think it should be \[\sum\limits_{k=1}^5k!\binom{5}{k}\] as we can permute the different flags too...

  6. Michele_Laino
    • one year ago
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    yes! @pgpilot326

  7. anonymous
    • one year ago
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    "any number" so since 0 is a number I guess that means 0 flags hoisted is a signal.

  8. anonymous
    • one year ago
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    so like @Michele_Laino said with the addition of 5 C 0 to make 2^5.

  9. Michele_Laino
    • one year ago
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    I think that you are right @ganeshie8 :)

  10. anonymous
    • one year ago
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    the number for @ganeshie8 's expression is just 2^5 - 1 (doesn't include the 5 C 0 term).

  11. ganeshie8
    • one year ago
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    :) i still feel that they should have explicitly mentioned whether permutations are considered or not..

  12. Michele_Laino
    • one year ago
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    I think that the permutations are admitted, since the order of the flags is not specified!

  13. phi
    • one year ago
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    it make sense permutations are allowed e.g. black over red means something different from red over black

  14. mathmath333
    • one year ago
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    why u not considered \(^{5}P_{0}\)

  15. mathmath333
    • one year ago
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    by the way answer in book is 325

  16. anonymous
    • one year ago
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    actually, order does matter since they are stacked. thus it should \[\sum_{k=0}^{5}\left(\begin{matrix}5 \\ k\end{matrix}\right)\cdot k!= \sum_{k=0}^{5}\frac{ 5! }{ \left( 5-k \right)!\cdot \cancel{k!} }\cdot \cancel{k!} =\sum_{k=0}^{5} {_5}P_{k}\]

  17. mathmath333
    • one year ago
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    Does at least one flag is necessary for a signal .

  18. anonymous
    • one year ago
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    =5+20+60+240 so 5P0 doesn't count as a signal. So has to have at least 1 flag

  19. anonymous
    • one year ago
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    yes but red over green is another signal, different from green over red

  20. anonymous
    • one year ago
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    order is implied in the question... "How many differenrt signals can be made by hoisting 5 different coloured flags \(\underline{\text{one above the other}}\)"

  21. ganeshie8
    • one year ago
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