Question

In how many ways 10 books can be arranged on a shelf such that a particular pair of books shall never be together

Answer 1

Here we 've the 10 books lying on shelf let there be no pair of books be come together so that we remained 9 out of 10 books which might take place one by one so that also subtracting pair of books we've left 8 ways so that total no of ways in which no particular pair of books come together is 8×9!

Answer 2

Answer:

Ans= 9!×8

Step-by-step explanation:

Total number of ways in which we can arrange 10 books on a shelf

= 10P10 =10! (A)

Now we will find out total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will always be together.

We have a total of 10 books. If a particular pair of books must always be together, just tie these two books together and consider as a single book. Hence we can take total number of books as 9. These 9 books can be arranged in 9P9 =9! ways.

We had tied two books together. These books can be arranged among themselves in 2P2 =2! ways.

Hence, total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will always be together

=9!×2! (B)

From (A) and (B),

Total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will never be together

=10!-(9!×2!)= 10!-(9!×2)= (9!×10)-(9!×2)= 9!(10-2)=9!×8

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