NIMCET 2019 Question Paper with solutions

Here, it is mentioned that repetition of digits, is not allowed
Now,for any number of four digits to be divisible by 5,its unit place must be occupied by 5

5

So, the number of ways to fill last place = 1
and the number of ways to fill remaining two place${=}^5{\mathit{P}}_3$
∴ Required number $=1{\times }^5{\mathit{P}}_3$
$=1\times \frac{5!}{\left(5-3\right)!}=1\times \frac{5\times 4\times 3\times 2!}{2!}=60$