In how many ways $$n$$ books can be arranged in a row so that two specified books are not together
The total no: of arrangement in which all $$n$$ books can be arranged on the shelf without any condition is $$^{n}P_{n}=n!$$ ...... $$(i)$$ 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 =$$2(n-1)!$$ ........ $$(ii)$$ The no: of ways $$ n $$ nooks on a shelf so that two particular books are not together is $$(i)-(ii)$$ $$=n!-2(n-1)!$$ Answer  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.1) n! – (n – 2) ! 2) (n – 1)! – (n – 2) 3) n! – 2(n – 1)! 4) (n – 2) n! Answer: (3) n! – 2(n – 1)! Solution: Total number of arrangement of n books = n! Two specified books can be together in 2P2 = 2! = 2 ways Let, the two specified books, kept together, as one composite book. Rest of the (n−2) books from (n−1) books can be arranged on the shelf in = n−1Pn−1
= (n−1)! Hence the number of ways the two specified books are together = 2(n−1)! [By fundamental rule of multiplication]Therefore, the required number of ways = n!−2(n−1)!
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The number of ways in which n books can be arranged so that two particular books are not together is
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