## 2 Answers

### 7 Comments

@Shubham Aggarwal I Think first can be written as -->(a)*(bb)* . Here both asterisks are independent of each other and can be thought as n and m.

I)Set of all strings containing any number of ‘a’ s followed by an even number of ‘b’ s. R.E=(a)$^{*}$(bb)$^{*}$.

IV) Strings containing a ‘c’. R.E= (a+b)$^{*}$c(a+b)$^{*}$.

Both these languages are regular as regular expressions exist.

By default a language is infinite. Eg : {a$^{n}$} it’s a infinite language.So both the languages II and III are infinite and comparison has to be done to evaluate these and hence are not regular.

Answer: A

NOTE:

Every finite language is regular.

Infinite language + Comparison = Non-Regular.

Infinite language + No Comparison = Regular.

Edit: As nothing is mentioned about ‘c’ in option IV and there is a comma after y, So I think It’s a typo ‘c’ should also belongs to {a,b}$^{*}$. IV will be a complete language. Which is regular. R.E=(a+b)$^{*}$.