Question

Number of ways in which 15 different books can be arranged on a shelf so that two particular books shall not be together is?
(a) $14\times 15!$
(b) $13\times 14!$
(c) $14!\times 15!$
(d) ${{\left( 15 \right)}^{2}}$

Answer 1

First find the number of ways to arrange all the 15 books on the shelf without any conditions using the relation $15!$. Now, find the number of ways to arrange the books with the condition that the two particular books are always together, to do so consider the two particular books as one unit and find the number of arrangements. Subtract the number of arrangements obtained in the latter case from the former case to get the answer.

Complete step by step answer:
(1) Here we are provided with 15 different books which are to be arranged on a bookshelf. We are asked to find the number of ways of arrangement such that two particular books are not together.
Now, we know that n different things can be arranged in $n!$ ways, so we get,
$\Rightarrow$ Number of ways in which 15 books can be arranged without any conditions = $15!$
(2) Let us find the number of ways of arrangement such that the two particular books are always together. The particular books that will always be together can be considered as one unit, so we have 14 things to arrange on the shelf (13 different books and one unit of two books).
$\Rightarrow$ Number of ways in which 14 items can be arranged = $14!$
The two particular books can be arranged among themselves in $2!$ ways, so we have,
$\Rightarrow$ Number of arrangements such that the two particular books always remain together = $2!\times 14!$
Therefore the number of arrangements where the two particular books will not be together will be the difference of the two cases, so we get,
$\Rightarrow$ Number of arrangements such that the two particular books will not be together = $15!-2!\times 14!$
$\Rightarrow$ Number of arrangements such that the two particular books will not be together = $14!\times \left( 15-2 \right)$
$\therefore$ Number of arrangements such that the two particular books will not be together = $13\times 14!$

So, the correct answer is “Option b”.

Note: Note that in the above solution you do not need to calculate the number of ways to select the two particular books using the formula of combinations ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ because they are already selected and that is why the term ‘particular’ is mentioned in the question. Remember the formulas of permutations and combinations used for the arrangements and selection respectively of the provided articles.