For a number to be divisible by 5, the number should end with 0 or 5.

If the number ends with 0.

Since 5 has to be one of the numbers, 5 can be fixed in any of the 4 places in 4 ways.

We now have 8 digits. which can be arranged in 8 * 7 * 6 ways for the 3 remaining places.

Therefore the number of numbers ending with 0 = 4 * 8 * 7 * 6 = 1344


If the number ends with 5.

The first place can have any of the 8 digits (excluding 0 and 5), the second place can again have 8 digits (excluding the number in the 1st place but with 0 being a possibility). The 3rd and the 4th place can have 7 and 6 digits respectively.

The total numbers with 5 in the units place = 8 * 8 * 7 * 6 = 2688

The total such numbers = 1344 + 2688 = 4032

4032 = k * 4! = 24k

k = 4032 / 24 = 168


Option B

Arun Kumar
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numbers divisible by 5 will have either 5 or 0 in the end

1 st case with last digit being 5

remaining 4 places left to be filled

since digits cannot repeat and first digit cant be 0
1st place can be filled in 8 ways
2nd place in 8 ways counting 0 also
3rd place in 7 ways
4th place in 6 ways

Thus 1st case = 8*8*7*6

2nd case with last digit being 0 has one more criteria of having a 5 also compulsorily

So remaining 3 places to be filled
1st place : 8 ways
2nd place : 7 ways
3rd place : 6 ways
and this can be done with having 5 in any of the 4 places
So
2nd Case = 8*7*6*4

adding both cases

(8*8*7*6)+(8*7*6*4)
taking 8*7*6 common (8+4)
=8*7*6(12)

Given K(4!)
4!=24

Hence
8*7*3*2(12)
=8*7*3*(24)
=168(24)
=168*(4!)

Option B