Question

The number of ways in which n books can be arranged can be arranged on a shelf so that two particular books shall not be together is

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

Answer 1

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

The total number of arrangements in which all $n$ books can be arranged

on the shelf without any condition is-

$W_1={}^n{P}_n=n!.....\left(1\right)$

Number of ways in which the two particular books are together $={}^2{P}_2=2!=2$ ways.

Consider those 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 number of ways $={}^{n-1}{P}_{n-1}=(n-1)!$

Therefore,

the total number of ways on which the two particular books are together-

$W_2=2\times (n-1)!.....\left(2\right)$

Now,

Number of ways of $n$ books on a shelf so that two particular books are not together is $=W_1-W_2$

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

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

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

Hence the correct answer is $\left(A\right)(n-2)(n-1)!$.