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

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(B) The total number of ways to arrange $n$ books in a row is $n!$.If two specified books are always together,we can treat them as a single unit. This

Vedclass · Accepted Answer

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

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

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

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