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In how many ways can 9 examination papers can arranged so that the best and the worst papers are never together?

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9! – 2! 7!

9! – 2! 8!

9! – 8!

9! – 7!

answer is B.

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Detailed Solution

$9! - [1 u n i t + 7 u n i t s = 8 u n i t s] = 9! - 2! \times 8!$

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