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In the process of 5B/6B encoding, it is possible to have the following two kinds of data sequences: 1. 25=32 data sequences 2. 26=64 data sequences Therefore, this can left us with some unused code sequences that are 32 in number (=26 – 25). On the other hand, in the encoding process of 3B/4B, it is possible to another set of data sequences as follows: 1. 23=8 data sequences 2. 24=16 data sequences This type of encoding also leaves us with some unused code sequences that are 8 in number (=23-24). 2. 

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We have 64 invalid code sequences in 5b/6b encoding. A further explanation is below.

  • The procedure of converting fragmented 5-bit statistical properties in and out of predefined 6-bit representations of characters is considered as 5B/6B Encoding.
  • This generates symmetrical observed data with an equivalent amount of 0's as well as 1's to ensure clocking transformation synchronization for receivers electronics as well as an appropriate authority value upon that line.

Thus the above answer is correct.

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