**Solution:** Treat the 3 specific books as a single block, which can be arranged internally in $3!$ ways. The block plus the remaining 5 books make 6 objects, which can be arranged in $6!$ ways. The number of ways to select the 3 specific books is $\binom{8}{3}$. However, since the 3 books are fixed in the block, we only consider arrangements of the block and the other 5 books. Therefore, total arrangements = $5! \times \binom{5}{3} \times 3!$.