How many different signals can be made by using at least three different flags if there are 6 different flags from which to select?
thanks
nope
because it says "at least 3" not "exactly 3"
so how should it be?? :O
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
well that was embarrassing *whistle*
\[\left(\begin{matrix}6 \\ 3\end{matrix}\right)+\left(\begin{matrix}6 \\ 4\end{matrix}\right)+\left(\begin{matrix}6 \\ 5\end{matrix}\right)+\left(\begin{matrix}6 \\ 6\end{matrix}\right)\]
@kropot72 is assuming order does not count in his/her computation
\[\sum_{k=3}^{6} \,_6P_k\]
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
more succinctly put by zarkon, that calculation is the one i wrote above
no
\[\frac{6!}{3!}+\frac{6!}{2!}+\frac{6!}{1}+\frac{6!}{1}\] As per @Zarkon
yes
wow was i ever off!
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