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

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

sirm3d
 2 years 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
 2 years 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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