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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
 one year ago
Nodata Group Title
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
 one year ago

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joemath314159 Group TitleBest ResponseYou've already chosen the best response.0
That 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.
 one year ago

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

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

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
the 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.
 one year ago

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

Nodata Group TitleBest ResponseYou've already chosen the best response.1
thank you @sirm3d
 one year ago
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