Final answer:
The correct grammar for the language L = n ≥ 0 over the alphabet A = {#, 0, 1} is not fully presented among the options, but option C is the closest. A more accurate representation would be using regular expressions such as '(01)*#'. (option C)
Step-by-step explanation:
The question is asking for the correct grammar of a given language over the alphabet A = {#, 0, 1}. The language is defined as L = n ≥ 0, meaning that the language consists of n repetitions of the pattern '01' followed by a '#', where n is greater than or equal to zero.
The correct grammar that generates this language is not provided in the options because grammar usually demonstrates how strings in a language are derived using production rules. However, the closest option to a correct form would be option C. S→01S | # because it recursively generates '01' multiple times until the termination symbol '#' is added.
Option C partially reflects the regular expression that would define the language, which would be more accurately stated as '(01)*#', meaning the sequence '01' can be repeated zero or more times followed by a '#'.