Question

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

• A. $n! - (n - 2)!$
• B. $(n - 1)!(n - 2)$
• C. $n! - 2(n - 1)$
• D. $(n - 2)n!$

Answer 1

Answer: (B)

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

Answer 2

Total number of arrangements of $n$ books $6P_{5} \times 5\mspace{6mu}! = 86400$.

If two specified books always together then number of ways $= (n - 1)\mspace{6mu}!\mspace{6mu} \times 2$

Hence required number of ways $= n\mspace{6mu}! - (n - 1)\mspace{6mu}!\mspace{6mu} \times 2$

$= n(n - 1)\mspace{6mu}!\mspace{6mu} - (n - 1)\mspace{6mu} \times 2 = (n - 1)\mspace{6mu}!\mspace{6mu}(n - 2)$.