Question Number 200284 by depressiveshrek last updated on 16/Nov/23
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{set}\:{A}\:\mathrm{containing}\:{n} \\ $$$$\mathrm{elements},\:\mid\mathcal{P}\left({A}\right)\mid=\mathrm{2}^{{n}} . \\ $$
Answered by AST last updated on 16/Nov/23
$${C}_{\mathrm{0}} ^{{n}} +^{{n}} {C}_{\mathrm{1}} +^{{n}} {C}_{\mathrm{2}} +…+^{{n}} {C}_{{n}} =\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =\mathrm{2}^{{n}} =\mid\mathcal{P}\left({A}\right)\mid \\ $$$${since}\:\mid\mathcal{P}\left({A}\right)\mid={Total}\:{number}\:{of}\:{subsets}\:{with}\: \\ $$$${cardinality}\:{of}\:\mathrm{0}\:{to}\:{n}. \\ $$