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

$n!–2(n–1)!$

Explanation for correct option:

Finding the number of ways$n$ books can be arranged in a row so that two specified books are not together

The number of arrangements such that all $n$books can be arranged in a row without any condition $$\begin{gathered} ={}^n{P}_n \\ =n!......\left(i\right) \end{gathered}$$

The number of ways to arrange the two books together $={}^{2}P_2$

$$\begin{gathered} =\frac{2!}{(2-2)!} \\ =2[\therefore 0!=1] \end{gathered}$$

Assume these two books are kept together as a single book.

Then we have the rest of the $(n-2)$ books from $(n-1)$ books which are to be arranged in a row.

Then the number of ways of arrangements of it $$\begin{gathered} ={}^{n-1}{P}_{n-1} \\ =(n-1)! \end{gathered}$$

Hence by the fundamental principle, the number of ways in which the two particular books are arranged together $=2(n-1)!........\left(ii\right)$

Thus the number of ways to arrange $n$ books in a row so that two particular books are not together is

$=n!-2(n-1)![\text{equation}(i)-\text{equation}(ii)]$

Hence, option (A) is the correct answer.