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
$$( n-2)n!$$
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:
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)!
Was this answer helpful?
4 (5)
Thank you. Your Feedback will Help us Serve you better.