Question

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

• A. $n!-(n-2)!$
• B. $(n-1)!-(n-2)$
• C. $n!-2(n-1)!$
• D. $(n-2)n!$

Answer 1

The correct option is C $n!-2(n-1)!$

The total no: of arrangement in which all $n$ books can be arranged on the shelf without any condition is

${}^n{P}_n=n!$ ...... $\left(i\right)$

The books can be together in ${}^2{P}_2=2!=2$ ways.

Consider these two books which are kept together as one composite book and with the rest of the $(n-2)$ books from$(n-1)$ books which are to be arranged on the shelf then the no: of ways=${}^{n-1}{P}_{n-1}=(n-1)!$

Hence by the fundamental principle , the no: of ways on which the two particular books are together =$2(n-1)!$ ........ $\left(ii\right)$

The no: of ways $n$ nooks on a shelf so that two particular books are not together is $\left(i\right)-\left(ii\right)$

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