SOLUTION: Among the 8! permutations of the digits 1,2,,,,,8 consider those arrangements which have the following property if you take any 5 consecutive ppositions the product of the digits

It's not any of those answers.

 1 2 3 4 5 6 7 8
 X X X X X _ _ _  <- the product of the digits where the X's are must have a 5.
 _ X X X X X _ _  <- the product of the digits where the X's are must have a 5.    
 _ _ X X X X X _  <- the product of the digits where the X's are must have a 5.
 _ _ _ X X X X X  <- the product of the digits where the X's are must have a 5.

The only one of the digits 1-8 that can cause a product
to be divisible by 5 is the number 5 itself.  So the 5 
must be in position 4 or 5 in order for you to be 
able to take any 5 consecutive positions and the product 
of the digits in those positions will be divisible by 5.  

So 
We can choose the position for the 5 in any of 2 ways.
We can choose the position for the 1 in any of the 7 remaining ways.
We can choose the position for the 2 in any of the 6 remaining ways.
We can choose the position for the 3 in any of the 5 remaining ways.
We can choose the position for the 4 in any of the 4 remaining ways.
We can choose the position for the 6 in any of the 3 remaining ways.
We can choose the position for the 7 in either of the 2 remaining ways.
We can choose the position for the 8 only the 1 remaining way.

Answer 2×7×6×5×4×3×2×1 2×7! = 10080.

Edwin