Question

How many 4 digit numbers are there, without repetition of digits, which are divisible by 5?

Answer 1

We know that, a number is divisible by $5$ if and only if it has $0$ or $5$ at its units place.

Hence, there are two cases:

1st case: 4 digit Number ending with 0.

$\underset{–}{}\underset{–}{}\underset{–}{}\underset{–}{0}$

By the method of permutation and combination,

The unit place can be filled with only $1$ number, that's $0$.

The thousands place can be filled with remaining $9$ numbers.

The hundreds place can be filled with remaining $8$ numbers.

The tens place can be filled with remaining $7$ numbers.

So, total number of four digit numbers, without repetition, having $0$ at their units place are $9\times 8\times 7\times 1=504$.

2nd case: 4 digit Number ending with 5.

$\underset{–}{}\underset{–}{}\underset{–}{}\underset{–}{5}$

The unit place can be filled with only $1$ number. that's $5$.

The thousands place cannot have $0$ and hence, can be filled with remaining $8$ numbers.

The hundreds place can be filled with remaining $8$ numbers. [This can have $0$]

The tens place can be filled with remaining $7$ numbers.

So total number of four digit numbers, without repetition, having $5$ at their units place are $8\times 8\times 7\times 1=448$.

Hence, total number of four digit numbers, without repetitions, which are divisible by $5$ are $504+448=952$.