In how many ways can $7$ different books be arranged on a shelf such that two specific books are never together?

Solution

1. Total arrangements of $7$ books is $7! = 5040$.
2. Treating the two specific books as one unit, there are $6!\times 2!$ arrangements where they are together.
3. $$\begin{array}{rl}6!\times 2& =720\times 2\\ & =1440\end{array}$$.
4. Required ways = $5040 - 1440 = 3600$.
5. Hence the answer is (C).