The string $1101$ does not belong to the set represented by
Only (a) and (b) can generate $1101$.
In (c) after $11$, we can not have $01$ and so $1101$ cannot be generated.
In (d) Every $11$ followed by $0$ and no single occurrence of $1$ is possible. So it cannot generate $1101$ or $11011$.
I think option (c) is more appropriate to choose from, because even option (d) can generate 11011 or 110110 in which 1101 is a substring.
yeahh. I think D is not allowed because let (00+(11)*0)* can be written as (a+b)*
and we know 'bb' is allowed in (a+b)* so in (00+(11)*0)* 110110110... is allowed. now see bold substring is 1101.
option d is quite inviting to choose it as we can have only even number of 1's with this regular set where as our original string contains 3 1's
from option c and d we can't generate string 1101.
therefore both c and d are correct