Product of the digits occupying the five consecutive positions is divisible by $5$. Find the number of permutations of $1,2,3,4,5,6,7,8$ taken all at a time without repetition in which the product of the digits occupying the five consecutive positions is divisible by $5$. In this question,i think $5$ is to be fixed at a place and rest of the digits is to be permuted.But i can not conclude the answer.Please help me.

Since the product of any 5 consecutive positions must be divisible by 5, 5 must either occupy position 4 or position 5, and the rest permuted around it, thus $2\cdot7!$ permutations