Question

Among the $8!$ permutations of the digits $1,2,\mathrm{3.........}8,$ 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 arrangement is equal to

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

Answer 1

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

We have $8$ positions the things is that the digit must be in first $5$ position or last $5$ position.

So, $5$ must be fixed at either the $4th$ place or $5th$ place and after that $7!$ combination of $7$ digits are possible.

$\therefore$ No. of ways $=2\times 7!$

Hence, the answer is $2(7!).$