In a city,no two persons have an identical se | vedclass

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(C) Each person has $32$ potential positions for teeth.For each position,there are $2$ possibilities: either a tooth is present or it is absent.Since

Vedclass · Accepted Answer

$2^{32} - 1$

Vedclass · Answer

(C) Each person has $32$ potential positions for teeth.For each position,there are $2$ possibilities: either a tooth is present or it is absent.Since there are $32$ such positions,the total number of possible combinations of teeth is $2 	imes 2 	imes \dots 	imes 2$ ($32$ times),which equals $2^{32}$.However,the problem states that there is no person without a tooth,meaning the case where all $32$ positions are empty must be excluded.Therefore,the maximum population of the city is $2^{32} - 1$.

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