What is the Cardinality of the Power set of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}?

harpazo Pure Mathematics Banned Oct 3, 2018 #1 Mar 20, 2018 5,857 361 83 NYC Oct 3, 2018 #1 What is the Cardinality of the Power set of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}?

MarkFL La Villa Strangiato Staff member Administrator Moderator Math Helper Oct 6, 2018 #2 Jan 25, 2018 3,463 4,254 113 St. Augustine Oct 6, 2018 #2 \(\displaystyle P=2^{10}=1024\) Suppose a set has a cardinality of \(n\). Then obviously, the number of subsets \(P\) that can be formed, is given by: \(\displaystyle P=\sum_{k=0}^n\left({n \choose k}\right)=(1+1)^n=2^n\) Likes: anemone and harpazo

\(\displaystyle P=2^{10}=1024\) Suppose a set has a cardinality of \(n\). Then obviously, the number of subsets \(P\) that can be formed, is given by: \(\displaystyle P=\sum_{k=0}^n\left({n \choose k}\right)=(1+1)^n=2^n\)

harpazo Pure Mathematics Banned Oct 6, 2018 #3 Mar 20, 2018 5,857 361 83 NYC Oct 6, 2018 #3 Interesting.

MarkFL La Villa Strangiato Staff member Administrator Moderator Math Helper Oct 6, 2018 #4 Jan 25, 2018 3,463 4,254 113 St. Augustine Oct 6, 2018 #4 The Binomial Theorem is quite powerful. Likes: anemone

harpazo Pure Mathematics Banned Oct 7, 2018 #5 Mar 20, 2018 5,857 361 83 NYC Oct 7, 2018 #5 Yes, it is....