mathmath333
  • mathmath333
Counting question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{How many differenrt signals can be made by hoisting }\hspace{.33em}\\~\\ & \normalsize \text{5 different coloured flags one above the other ,when }\hspace{.33em}\\~\\ & \normalsize \text{any number of them may be hoisted at a time }\hspace{.33em}\\~\\ & a.)\ 2^{5} \hspace{.33em}\\~\\ & b.)\ ^{5}P_{5} \hspace{.33em}\\~\\ & c.)\ 325 \hspace{.33em}\\~\\ & d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
is hoisting 0 flags considered a signal or no?
Michele_Laino
  • Michele_Laino
I think that the requested number is given by the subsequent computation: \[\Large \left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 5 \end{array}} \right) = {\left( {1 + 1} \right)^5}\]

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anonymous
  • anonymous
@Michele_Laino you forgot 5C0
ganeshie8
  • ganeshie8
i think it should be \[\sum\limits_{k=1}^5k!\binom{5}{k}\] as we can permute the different flags too...
Michele_Laino
  • Michele_Laino
yes! @pgpilot326
anonymous
  • anonymous
"any number" so since 0 is a number I guess that means 0 flags hoisted is a signal.
anonymous
  • anonymous
so like @Michele_Laino said with the addition of 5 C 0 to make 2^5.
Michele_Laino
  • Michele_Laino
I think that you are right @ganeshie8 :)
anonymous
  • anonymous
the number for @ganeshie8 's expression is just 2^5 - 1 (doesn't include the 5 C 0 term).
ganeshie8
  • ganeshie8
:) i still feel that they should have explicitly mentioned whether permutations are considered or not..
Michele_Laino
  • Michele_Laino
I think that the permutations are admitted, since the order of the flags is not specified!
phi
  • phi
it make sense permutations are allowed e.g. black over red means something different from red over black
mathmath333
  • mathmath333
why u not considered \(^{5}P_{0}\)
mathmath333
  • mathmath333
by the way answer in book is 325
anonymous
  • anonymous
actually, order does matter since they are stacked. thus it should \[\sum_{k=0}^{5}\left(\begin{matrix}5 \\ k\end{matrix}\right)\cdot k!= \sum_{k=0}^{5}\frac{ 5! }{ \left( 5-k \right)!\cdot \cancel{k!} }\cdot \cancel{k!} =\sum_{k=0}^{5} {_5}P_{k}\]
mathmath333
  • mathmath333
Does at least one flag is necessary for a signal .
anonymous
  • anonymous
=5+20+60+240 so 5P0 doesn't count as a signal. So has to have at least 1 flag
anonymous
  • anonymous
yes but red over green is another signal, different from green over red
anonymous
  • anonymous
order is implied in the question... "How many differenrt signals can be made by hoisting 5 different coloured flags \(\underline{\text{one above the other}}\)"
ganeshie8
  • ganeshie8
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