Answer:(A)
Explanation: The hamming distance of strings S1 and S2 can be any value between 0 and n. Hamming distance is the number of different bits between two strings. For a given string and d distance, there can be C(n,d) strings having d as the hamming distance as from n bits any d need to be chosen. For any string S1, there are – C(n,0) strings with 0 hamming distance C(n,1) strings with 1 hamming distance C(n,2) strings with 2 hamming distance . . C(n,n) strings with n hamming distance Total strings = C(n,0) + C(n,1) + … + C(n,n) =
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