Let $L_1$ be the language represented by the regular expression $b^*ab^*(ab^*ab^*)^*$ and $L_2 = \{ w \in (a + b)^* \mid |w| \le 4 \}$, where $|w|$ denotes the length of string $w$. The number of strings in $L_2$ which are also in $L_1$ is _________

1. Language $L_1$: The regular expression $b^*ab^*(ab^*ab^*)^*$ represents all strings over $\{a, b\}$ that contain an odd number of $a$'s.

2. Language $L_2$: $L_2$ contains all strings over $\{a, b\}$ of length at most 4 (i.e., $|w| \le 4$).

3. Intersection $L_1 \cap L_2$: We need to count strings of length $|w| \le 4$ having an odd number of $a$'s (which means either 1 or 3 occurrences of $a$):

   • Length 0 ($|w| = 0$): 0 strings.

   • Length 1 ($|w| = 1$): Exactly 1 '$a$' $\implies \binom{1}{1} = 1$ string (a).

   • Length 2 ($|w| = 2$): Exactly 1 '$a$' $\implies \binom{2}{1} = 2$ strings (ab, ba).

   • Length 3 ($|w| = 3$):

     • 1 '$a$': $\binom{3}{1} = 3$ strings

     • 3 '$a$'s: $\binom{3}{3} = 1$ string

     • Total = $3 + 1 = 4$ strings.

   • Length 4 ($|w| = 4$):

     • 1 '$a$': $\binom{4}{1} = 4$ strings

     • 3 '$a$'s: $\binom{4}{3} = 4$ strings

     • Total = $4 + 4 = 8$ strings.

4. Total Count:

   $$1 + 2 + 4 + 8 = 15$$

Thus, the number of strings in $L_2$ which are also in $L_1$ is 15.