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How many five-digit integers can be possible?[NOTE:- not  positive integers]
in Mathematical Logic
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it will be 18*10= 180000

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Each of the 4 places barring the unit's place will have 10 choices nd the unit's place will have 9 (excluding 0) choices

So, total 5-digit numbers possible = 9*10*10*10*10 = 9*104

and as we know number of positive n-digit integers = number of negative n-digit integers, 

So, total 5-digit integers possible = 2*9*104  = 18*104

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