The following finite state machine accepts all those binary strings in which the number of $1$’s and $0$’s are respectively:
divisible by $3$ and $2$
odd and even
even and odd
divisible by $2$ and $3$
Option (C) and option (D) are cancelled out clearly because with $3$ $1$s we can reach the final state. There is an string where we can reach the final state by $6$ $1$'s now $6$ is nt odd but it is divisivble by $3$. Hence, option (A) is correct.
option d is 1 divisible by 2 and 0 is divisible 3 which is wrong.
No. of 0's % 2 = 0
no. of 1's % 3 =0
ε (Empty string) is accepted here. So option (B) & (C) are canclelled out !
No of 0's % 2 = 0 . So answer => A