\textbf{Divisibility rule of 4:} \\
\text{If the last two digits of a number are divisible by 4, then that number is divisible by 4.} \\[6pt]
\textbf{Permutation:} \\
\text{Permutation is defined as an arrangement of } r \text{ things that can be done out of total } n \text{ things.} \\
\text{This is denoted by } \ ^nP_r = \dfrac{n!}{(n - r)!} \\[8pt]
\textbf{Calculation:} \\[4pt]
\text{Given digits are } 1, 2, 3, 4, 5, 6 \\[4pt]
\text{As we know, a number is divisible by 4, the last two digits must be divisible by 4.} \\[4pt]
\text{So, number of possible last two digits will be } \{12, 16, 24, 32, 36, 52, 56, 64\} \\[4pt]
\text{Number of ways of selecting one of these numbers } = \ ^8C_1 = 8 \text{ ways.} \\[10pt]
\text{We have already taken the last two digits, therefore } n = 6 - 2 = 4. \\[4pt]
\text{So, number of digits to be filled is } 3 = \ ^4P_3 \\[4pt]
= 4 \times 3 \times 2 = 24 \\[10pt]
\text{Now, total number of 5-digit numbers which will be divisible by 4 is} \\[4pt]
= (\text{Number of ways of selecting first 3 digits}) \times (\text{Number of ways of selecting last 2 digits}) \\[4pt]
= 24 \times 8 = 192 \\[6pt]
\text{Hence, the total number of such 5-digit numbers is } 192.
