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Among 9! permutations of the digits 1,2,3,4....9. Consider those arrangements which have the property that we take any five consecutive positions, the product of the digits in those positions is divisible by 7, the number of such arrangement is ??

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    In order for the product to be divisible by $7$, one of the digits included must be ....? – 2017-02-15
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    According to me ... it should contain 7. – 2017-02-15
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    Where should the $7$ be for the property to hold? – 2017-02-15
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    Hint: you have nine positions if the seven is not in the first five positions the property fails, if it's not in the last five positions it also fails. So where should it be? – 2017-02-15

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7 must come in any of the middle position. So 8! ways

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    Sorry, I didn't get your logic and its wrong. – 2017-02-15
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    I think you think this is about any seven consecutive digits. It's not, it's about any five consecutive digits. – 2017-02-15
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    $7$ must come in the middle of **every permutation**, so only $8!$ ways in total. – 2017-02-15
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    yeah right! @barakmanos. the answer is right but how it comes out to be 8! ? – 2017-02-15
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    @abhishek: Put $7$ in the middle, and arrange the remaining $8$ digits in any possible manner. – 2017-02-15
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    Arthur u r right. I read it wrongly. Corrected – 2017-02-15