Question

In how many ways can $10$ books be arranged in a shelf such that a particular pair of books are never together?

• A. $9!\times 8$
• B. $9!$
• C. $9!\times 2!$
• D. $10!\times 2!$

Answer 1

The correct option is A $9!\times 8$

Number of books $=10$

Total number of ways $=10!$

Given pair of books to be together considering them as one

total sets $=9$

Number of ways of arranging them $=9!$ways

Those two books can be arranged in $2!$

Total ways that the pair of books is together

$\Rightarrow 9!\times 2!$

Number of ways that are not together

$\Rightarrow 10!-9!\times 2!\Rightarrow 9!(10-2)\Rightarrow 8\times 9!$