To solve this, we need to consider two aspects of the problem:

1. The books of the same subject must stay together:

   • This means that we can treat the books of each subject (English, Science, and Mathematics) as a single unit or "block."
   • Since there are 3 subjects, we can arrange these 3 blocks in $3!$ ways.

2. Arranging the books within each subject block:

   • The 3 English books can be arranged among themselves in $3!$ ways.
   • The 6 Science books can be arranged among themselves in $6!$ ways.
   • The 4 Mathematics books can be arranged among themselves in $4!$ ways.

Thus, the total number of ways to arrange the books is the product of these possibilities:

$3! \times 3! \times 6! \times 4!$

This matches option c. $3!6!4!3!$.

Answer: c. $3!6!4!3!$

Would you like further details or have any questions?

Here are 5 related questions:

1. What is the value of $3!$, $6!$, and $4!$?
2. How many ways can 5 different books be arranged on a shelf?
3. What if there were 2 more subjects with the same condition?
4. What if the books could be placed in any order without restriction?
5. How does this problem change if the books within the same subject can only be in alphabetical order?

Tip: Factorials grow very quickly as numbers get larger, so simplifying them where possible helps in calculations.