anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
That formula is right, but there is a much shorter formula for the number of elements in the power set of an n-element set. If write out the power sets for sets with 2, 3 and 4 elements, it should become clear what the shorter formula is.
anonymous
  • anonymous
Yeah I know. The formula is \(2^n\)
anonymous
  • anonymous
I'm ashamed this is actually the binomial theorem

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

sirm3d
  • sirm3d
the binomial expansion of \((x+y)^n\) is \[\Large (x+y)^n=\sum_{k=0}^n \left(\begin{matrix}n \\ k\end{matrix}\right)x^ky^{n-k}\] put \(x=y=1\) and you'll get the answer to your question.
anonymous
  • anonymous
\[\#\wp S=2^{\#S}= \sum_{k=0}^{n} {n\choose k}\]
anonymous
  • anonymous
thank you @sirm3d
sirm3d
  • sirm3d
YW

Looking for something else?

Not the answer you are looking for? Search for more explanations.