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Mathematics 9 Online
OpenStudy (anonymous):

How many different signals can be made by using at least three different flags if there are 6 different flags from which to select?

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

because it says "at least 3" not "exactly 3"

OpenStudy (lgbasallote):

so how should it be?? :O

OpenStudy (anonymous):

so it could be 3, 4, 5, or 6 flags used plus, it is not clear if order counts for this one i would imagine it would but i could be wrong

OpenStudy (lgbasallote):

well that was embarrassing *whistle*

OpenStudy (kropot72):

\[\left(\begin{matrix}6 \\ 3\end{matrix}\right)+\left(\begin{matrix}6 \\ 4\end{matrix}\right)+\left(\begin{matrix}6 \\ 5\end{matrix}\right)+\left(\begin{matrix}6 \\ 6\end{matrix}\right)\]

OpenStudy (anonymous):

@kropot72 is assuming order does not count in his/her computation

OpenStudy (zarkon):

\[\sum_{k=3}^{6} \,_6P_k\]

OpenStudy (anonymous):

and i don't know if that is correct or not, but i would assume that say Red, Green, Yellow is different from Green, Yellow, Red etc

OpenStudy (anonymous):

more succinctly put by zarkon, that calculation is the one i wrote above

OpenStudy (zarkon):

no

OpenStudy (kropot72):

\[\frac{6!}{3!}+\frac{6!}{2!}+\frac{6!}{1}+\frac{6!}{1}\] As per @Zarkon

OpenStudy (zarkon):

yes

OpenStudy (anonymous):

wow was i ever off!

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