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 one year ago
I was enumerating the elements of the power set of this set S:= {1,2,3,4,5} and I thought that the number of these elements could be obtained with this:
\[\#\wp S = 1 + \sum_{k=1}^n {n\choose k}
\]
where \(n=\#S\)
I saw that it holds for this set. But I'm not sure what if it could be applied to a different kind of set.
 one year ago
I was enumerating the elements of the power set of this set S:= {1,2,3,4,5} and I thought that the number of these elements could be obtained with this: \[\#\wp S = 1 + \sum_{k=1}^n {n\choose k} \] where \(n=\#S\) I saw that it holds for this set. But I'm not sure what if it could be applied to a different kind of set.

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joemath314159
 one year ago
Best ResponseYou've already chosen the best response.0That formula is right, but there is a much shorter formula for the number of elements in the power set of an nelement set. If write out the power sets for sets with 2, 3 and 4 elements, it should become clear what the shorter formula is.

Nodata
 one year ago
Best ResponseYou've already chosen the best response.1Yeah I know. The formula is \(2^n\)

Nodata
 one year ago
Best ResponseYou've already chosen the best response.1I'm ashamed this is actually the binomial theorem

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.1the binomial expansion of \((x+y)^n\) is \[\Large (x+y)^n=\sum_{k=0}^n \left(\begin{matrix}n \\ k\end{matrix}\right)x^ky^{nk}\] put \(x=y=1\) and you'll get the answer to your question.

Nodata
 one year ago
Best ResponseYou've already chosen the best response.1\[\#\wp S=2^{\#S}= \sum_{k=0}^{n} {n\choose k}\]
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