harpazo Pure Mathematics Oct 4, 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 Math Helper 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: \(\displaystyle P=\sum_{k=0}^n\left({n \choose k}\right)=(1+1)^n=2^n\)

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