In how many ways n books can be arranged in a | vedclass

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(B) The total number of ways to arrange $n$ books in a row is $n!$.To find the number of ways where two specified books are together,we treat them as

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$(n - 1)! (n - 2)$

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(B) The total number of ways to arrange $n$ books in a row is $n!$.To find the number of ways where two specified books are together,we treat them as a single unit. This leaves us with $(n - 1)$ units to arrange,which can be done in $(n - 1)!$ ways. The two specified books can be arranged among themselves in $2! = 2$ ways.Thus,the number of ways the two specified books are together is $2 	imes (n - 1)!$.The number of ways the two specified books are $NOT$ together is the total arrangements minus the arrangements where they are together:$= n! - 2(n - 1)!$$= n(n - 1)! - 2(n - 1)!$$= (n - 1)! (n - 2)$.

In how many ways $n$ books can be arranged in a row so that two specified books are not together?

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