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An n- bit string is n- digit "word" , each digit of which is either 0 or 1.
Find the no. of n- bit strings containing both the digit zero and one.
Dear Sanchit
Each position of n digit string can be filled by 2 ways either 0 or 1
so total number os such word is
=2*2*2*2*2 .......n times
=2n
but it include that case also which has 0 at all position and 1 at all position
So total number of desired way =2n-2
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