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Question

In how many ways $$n$$ books can be arranged in a row so that two specified books are not together


A
n!(n2)!
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B
(n1)!(n2)
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C
n!2(n1)!
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D
(n2)n!
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Solution

The correct option is C $$n!- 2( n-1)!$$
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 $$(i)-(ii)$$
$$=n!-2(n-1)!$$

Maths

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