We know, any number to be divisible by 4, we must have the last 2 digits of the number divisible by 4.

So, the whole four-digit number will be divisible by 4 if the last 2 digits are: 12, 32, 52, 24.

Case-1: When, we have 2 in ones’ place:

In this case, we can fill tens’ place with 1 ,3 or 5 i.e. we have 3 ways to fill the tens’ place.

Hundreds’ place can be filled with any 2 of the remaining 3 digits [2 digits are used to fill the ones’ and tens’ place among the given 5 digits], in ³P₂ = 6 ways.

So, number of four-digit numbers that can be formed in this case

= 18.

Case-2: When, we have 4 in ones’ place:

In this case, we can fill tens’ place with 2 i.e. tens’ place can be filled only in 1 way.

Hundreds’ place can be filled with any 2 of the remaining 3 digits [2 digits are used to fill the ones’ and tens’ place among the given 5 digits], in ³P₂ = 6 ways.

So, number of four-digit numbers that can be formed in this case = 6.

As, the two cases are independent so, the total number of four-digit numbers that can be formed of the digits 1, 2, 3, 4 and 5 and divisible by 4 is = 18 + 6

= 24.