Question

In how many ways 10 books can be arranged on a shelf so that a particular pair of book shall be
a) always together ?
b) never together?

Answer 1

Given that 10 books can be arranged on a shelf is 10 ! ways

a) Pair of books always together i.e 2 books will take 1 unit and remaining 8 books on 8 units then total number of units = 1 + 8 = 9.

These 9 can be arranged in 9! ways and two Pair of books can be arranged in 2! ways.

∴ hence the total number of arrangements = 9! x 2! ways.

b)
First 8 books can be arranged in 8! ways and there are 9 gaps and 2 books can be

arranged in 9P2 ways.

∴ hence the total number of arrangements = 8! x 9P2 ways.

Answer 2

Answer:

Ans= (a)9!*2!    (b)9!×8   Mark me Brainliest

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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