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mathmath333
 one year ago
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mathmath333
 one year ago
Counting question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{How many differenrt signals can be made by hoisting }\hspace{.33em}\\~\\ & \normalsize \text{5 different coloured flags one above the other ,when }\hspace{.33em}\\~\\ & \normalsize \text{any number of them may be hoisted at a time }\hspace{.33em}\\~\\ & a.)\ 2^{5} \hspace{.33em}\\~\\ & b.)\ ^{5}P_{5} \hspace{.33em}\\~\\ & c.)\ 325 \hspace{.33em}\\~\\ & d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is hoisting 0 flags considered a signal or no?

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino you forgot 5C0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i think it should be \[\sum\limits_{k=1}^5k!\binom{5}{k}\] as we can permute the different flags too...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! @pgpilot326

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"any number" so since 0 is a number I guess that means 0 flags hoisted is a signal.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so like @Michele_Laino said with the addition of 5 C 0 to make 2^5.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that you are right @ganeshie8 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the number for @ganeshie8 's expression is just 2^5  1 (doesn't include the 5 C 0 term).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2:) i still feel that they should have explicitly mentioned whether permutations are considered or not..

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that the permutations are admitted, since the order of the flags is not specified!

phi
 one year ago
Best ResponseYou've already chosen the best response.0it make sense permutations are allowed e.g. black over red means something different from red over black

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0why u not considered \(^{5}P_{0}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0by the way answer in book is 325

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually, order does matter since they are stacked. thus it should \[\sum_{k=0}^{5}\left(\begin{matrix}5 \\ k\end{matrix}\right)\cdot k!= \sum_{k=0}^{5}\frac{ 5! }{ \left( 5k \right)!\cdot \cancel{k!} }\cdot \cancel{k!} =\sum_{k=0}^{5} {_5}P_{k}\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0Does at least one flag is necessary for a signal .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0=5+20+60+240 so 5P0 doesn't count as a signal. So has to have at least 1 flag

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but red over green is another signal, different from green over red

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0order 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440617925160:dw