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Prove-that-for-any-set-A-containing-n-elements-P-A-2-n-




Question Number 200284 by depressiveshrek last updated on 16/Nov/23
Prove that for any set A containing n  elements, ∣P(A)∣=2^n .
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{set}\:{A}\:\mathrm{containing}\:{n} \\ $$$$\mathrm{elements},\:\mid\mathcal{P}\left({A}\right)\mid=\mathrm{2}^{{n}} . \\ $$
Answered by AST last updated on 16/Nov/23
C_0 ^n +^n C_1 +^n C_2 +...+^n C_n =(1+1)^n =2^n =∣P(A)∣  since ∣P(A)∣=Total number of subsets with   cardinality of 0 to n.
$${C}_{\mathrm{0}} ^{{n}} +^{{n}} {C}_{\mathrm{1}} +^{{n}} {C}_{\mathrm{2}} +…+^{{n}} {C}_{{n}} =\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =\mathrm{2}^{{n}} =\mid\mathcal{P}\left({A}\right)\mid \\ $$$${since}\:\mid\mathcal{P}\left({A}\right)\mid={Total}\:{number}\:{of}\:{subsets}\:{with}\: \\ $$$${cardinality}\:{of}\:\mathrm{0}\:{to}\:{n}. \\ $$

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