number of ways we can arrange 1,2,3,4,...8 such that the product of any five consecutive position is divisible by 5

Hint: 

First find out the possible ways of product of five numbers havinng last digit 0 or 5 .

u will have to check this first.

then find the consecutiveness order .

If 5 is placed in the 4^th or 5^th place, then the product of any 5 consecutively placed numbers will be divisible by 5. So, for 5, we have two choices of placement, and corresponding to each, the other digits can be permuted in 7! ways.

 

Hence, total number of ways = 2 X 7! = 10080