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Consider a binary channel code in which each codeword has a fixed length of 5 bits. The Hamming distance between any pair of distinct codewords in this code is at least 2. The maximum number of codewords such a code can contain is _________.

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for any code C(n,k,d), d <= n-k+1

      where, d=minimum hamming distance.

                  k=dataword length

                 n=codeword length

d=2,n=5  =>  2<=5-k+1

solving it, k=4

so, maximum number of codewords possible is $2^{4}$=16.

note: this equation, i've written above is singleton bound equation. i don't think it's important or is even in our syllabus. nevertheless, if you want, you can refer here. https://www.cse.iitm.ac.in/~jayalal/teaching/CS6845/2012/lecture-B04.pdf

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Consider 5 bit grey code where each consecutive code word differs by one bit.Considering this code , we have to choose maximum number of codes among these 32 codes ,such that no two are consecutive, hence if we choose 17 or greater number of codewords ,atleast two words are consecutive. So we can choose 16 codewords leaving one space between each of them.

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