Question

In how many ways n books can be arranged so that 2 particular books are not together.Plz solve this Q step by step

Answer 1

leaving the two particular books aside, we first arrange( n-2) books in

$(n - 2) \: factorial \: ways$

now there are( n-1) places for two particular books inorder to arrange in such a way that they are not together. it can be done in (n-1)(n-2) ways.

required number=(n-2)![(n-1)(n-2)]

Answer 2

The number of ways n books can be arranged so that 2 particular books are not together is n ! - 2(n - 1) !

Solution:

Given that,

We have to find the number of ways n books can be arranged so that 2 particular books are not together

To arrange n books, the number of ways = n !

Consider that particular books are always together and is treated as 1 object

So there are (n - 1) objects that can be arranged in (n - 1)! ways

Two books can be arranged between themselves in 2 ways

Number of ways two books are always together is: $2 (n - 1)!$

Therefore, number of ways in two books are not together is:

$n! - 2(n-1)!$

Thus the number of ways n books can be arranged so that 2 particular books are not together is n ! - 2(n - 1) !

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