Discussion
The following CFG S → aB | bA, A → a | as | bAA, B → b | bs | aBB generates strings of terminals that have
Answer: C
The grammar generates strings with equal numbers of a's and b's. A generates the strings that have one more a than b's, B generates those with one more b than a's, and S starts with a followed by a string with one more b, or b followed by a string with one more a, so the totals are always equal. A check: S gives aB which gives ab, and bA gives ba, each with one a and one b. So the strings have an equal number of a's and b's, option (C). The options on odd, even and unequal counts are therefore wrong.
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