Question

In how many ways n books can be arranged in a row so that two specified books are not together
A) \[n!-\left( {n-2} \right)!\]
B) \[\left( {n-1} \right)! - \left( {n-2} \right)\]
C) \[n!-2\left( {n-1} \right)!\]
D) \[\left( {n-2} \right)n!\]

Answer 1

Hint:
First, find the number of ways in which n books can be arranged in a row.
Then, find the number of ways that any 2 books are kept together.
Finally, to get the number of ways n books can be arranged in a row so that two specified books are not together can be given by subtracting the number of ways in which 2 books can be arranged together from the total number of ways n books can be arranged.

Complete step by step solution:
Firstly, the number of ways n books can be arranged in different ways without any condition is, ${}^n{P_n} = \dfrac{{n!}}{{\left( {n - n} \right)!}} = \dfrac{{n!}}{{0!}} = n!$ … (1)
Now, 2 books can be together in ${}^2{P_2} = \dfrac{{2!}}{{\left( {2 - 2} \right)!}} = \dfrac{{2!}}{{0!}} = 2! = 2$ different ways.
Let us consider these 2 books as one whole composite book and kept together with the remaining \[\left( {n-2} \right)\] books.
Now, the remaining \[\left( {n - 1} \right)\] books can be arranged in ${}^{n - 1}{P_{n - 1}} = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 1 - n + 1} \right)!}} = \dfrac{{\left( {n - 1} \right)!}}{{0!}} = \left( {n - 1} \right)!$ .
Then, by the fundamental principle, the number of ways in which 2 books can be arranged together becomes \[2\left( {n-1} \right)!\] .
Now, the number of ways n books can be arranged in a row so that two specified books are not together is given by subtracting the number of ways in which 2 books can be arranged together i.e. \[2\left( {n-1} \right)!\] from total number of ways n books can be arranged i.e. \[n!\] .

So, the number of ways n books can be arranged in a row so that two specified books are not together $ = n! - 2\left( {n - 1} \right)!$.

Note:
Permutation is generally used in the lists where order of the objects matter. Here, specific two books cannot be arranged together. So, we have to use permutation.
Combination is generally used for groups where order of the object does not matter.