Permutations n combinations
Probability
Level
2
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 digits in those positions is divisible by 5. The number of such arrangements is :
7!
2*7!
None of these
8*7!
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1 solution
We require the number '5' to always be in the 5 consecutive positions because then the product of those digits will always be divisible by '5'. With 8 digits 1,..,8, there are two digit positions that allows this. The 4th digit or 5th digit (counting from left or right). There are 2 ways to pick 5 in this position and once fixed, 7! ways to permute the remainder digits.