In how many ways can 9 examination papers be | vedclass

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(B) The total number of ways to arrange $9$ papers is $9!$. To find the number of ways where the best and worst papers are together,we treat them as a

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$9! - 2! 	imes 8!$

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(B) The total number of ways to arrange $9$ papers is $9!$. To find the number of ways where the best and worst papers are together,we treat them as a single unit. This leaves us with $8$ units (the combined pair plus the other $7$ papers),which can be arranged in $8!$ ways. The best and worst papers can be arranged within their unit in $2!$ ways. So,the number of ways they are together is $2! 	imes 8!$. The number of ways they are never together is the total arrangements minus the arrangements where they are together: $9! - 2! 	imes 8!$.

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

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