Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Permutations and Combinations

Solution:

image
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 $2 \times 7!$.