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 to

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2.(7!)

7C4

none of these

answer is B.

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Detailed Solution

Let the arrangement be x1 x2 x3 x4 x5 x6 x7 x8. Clearly 5 should occupy the position x4 or x5. Thus, required number of arrangements is 2(7!).

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