Question

Counting question

Answer 1

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

Answer 2

is hoisting 0 flags considered a signal or no?

Answer 3

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

Answer 4

@Michele_Laino you forgot 5C0

Answer 5

i think it should be
\[\sum\limits_{k=1}^5k!\binom{5}{k}\]

as we can permute the different flags too...

Answer 6

yes! @pgpilot326

Answer 7

"any number" so since 0 is a number I guess that means 0 flags hoisted is a signal.

Answer 8

so like @Michele_Laino said with the addition of 5 C 0 to make 2^5.

Answer 9

I think that you are right @ganeshie8 :)

Answer 10

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

Answer 11

:) i still feel that they should have explicitly mentioned whether permutations are considered or not..

Answer 12

I think that the permutations are admitted, since the order of the flags is not specified!

Answer 13

it make sense permutations are allowed
e.g. black over red means something different from red over black

Answer 14

why u not considered \(^{5}P_{0}\)

Answer 15

by the way answer in book is 325

Answer 16

actually, order does matter since they are stacked.

thus it should \[\sum_{k=0}^{5}\left(\begin{matrix}5 \\ k\end{matrix}\right)\cdot k!= \sum_{k=0}^{5}\frac{ 5! }{ \left( 5-k \right)!\cdot \cancel{k!} }\cdot \cancel{k!} =\sum_{k=0}^{5} {_5}P_{k}\]

Answer 17

Does at least one flag is necessary for a signal .

Answer 18

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

Answer 19

yes but red over green is another signal, different from green over red

Answer 20

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