Question

Among the $8!$ permutations of the digits $1,2,3,...,8$, consider those arrangements which have the following property. If we take any five consecutive positions, the product of the digits in these positions is divisible by $5$. The number of such arrangements is equal is

• A. $7!$
• B. $2.(7!)$
• C. ${}^7{C}_4$
• D. None of these

Answer 1

The correct option is B $2.(7!)$

Let the arrangement be $x_1{x}_2{x}_3{x}_4{x}_5{x}_6{x}_7{x}_8$ clearly $5$ should occupy the position $x_4$ or $x_5$. Thus required number is $2(7!)$.