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

Question

Among 8! permutations of the digits 1, 2, 3, ..., 8, consider those arrangements which have the following property: if you take any five consecutive positions, the product of the digits in those positions is divisible by 5. The number of such arrangements is (A) 7 ! (B)  2⋅ 7 ! (C) 8 ⋅ 7! (D)  (7 !)2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/633813deb60cdbf6f12456353aaf343b-.png" /> According to the given arrangements, 5 can be put in either of the two places shown in the figure out of given 8 positions. Hence total number of such arrangements will be 2 × 7!.

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

$7!$

$2 \cdot 7!$

$8 \cdot 7!$

$(7!)^{2}$

According to the given arrangements, $5$ can be put in either of the two places shown in the figure out of given $8$ positions. Hence total number of such arrangements will be $2 \times 7!$ .

Q. Among 8! permutations of the digits 1,2,3,...,8, consider those arrangements which have the following property: if you take any five consecutive positions, the product of the digits in those positions is divisible by 5. The number of such arrangements is

7!

2⋅7!

8⋅7!

(7!)2