## mathmath333 one year ago Counting question

1. 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}}

2. anonymous

is hoisting 0 flags considered a signal or no?

3. Michele_Laino

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

4. anonymous

@Michele_Laino you forgot 5C0

5. ganeshie8

i think it should be $\sum\limits_{k=1}^5k!\binom{5}{k}$ as we can permute the different flags too...

6. Michele_Laino

yes! @pgpilot326

7. anonymous

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

8. anonymous

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

9. Michele_Laino

I think that you are right @ganeshie8 :)

10. anonymous

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

11. ganeshie8

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

12. Michele_Laino

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

13. phi

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

14. mathmath333

why u not considered $$^{5}P_{0}$$

15. mathmath333

by the way answer in book is 325

16. anonymous

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

17. mathmath333

Does at least one flag is necessary for a signal .

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. ganeshie8

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