How many arrangements of $4$ math, $6$ physics, and $2$ chemistry books are possible if the books in a particular subject must stand together? Here is the question: In how many ways can $4$ different math books, $6$ different physics books, and $2$ different chemistry books be arranged on a shelf if books in a particular subject must stand together?
The solution is the source (https://socratic.org/questions/different-4-math-books-6-different-physics-books-2-different-chemistry-books-are).
I know that permutations are used when order matters and combinations are used when order does not matter. They are both used when you do not allow repetition and replacement.
My question is why the solution uses permutations in this problem. Should we be using combinations as order does not matter? Order does not matter as it does not matter the order of the math books within the math group, same as the other subjects. Also, the order of the subjects does not matter. It does not matter if it is math first, then the physics, and then chemistry.
How can we change the wording of this question to make us use combinations?
I am still confused on when to use combinations and when to use permutations in a word problem.
Thank you so much.
 A: Because the math books are specified as different, we care about the order of the four of them on the shelf.  There are $4!$ ways to arrange the math books among themselves.  We also care about whether the math books are at the left end, in the middle, or on the right end.  There are $3!$ orders for the three subjects.
The whole problem is written to use permutations, using the word different for the books in each subject and using distinct subjects.  If we said the math section was four copies of the same book we would use combinations and find only one arrangement for the math books among themselves.  Alternately we could say that only some of the books were to be shelved and ask how many sets of books there were.  This would remove the interest in the order the books were on the shelf.  If only two math books were shelved, there would be $4 \choose 2$ ways to choose which $2$.
A: If the order of the books in a specific subject does not matter, then we can treat each group of book as a "block", and simply count how many ways there are of permuting three "blocks", which is $3!=6.$ However, if the order of the books in a specific subject does matter, we need to multiply by the number of ways of permuting each subject group. Then, the answer would be
$$3!\times 6!\times 2!\times 4!=207360.$$
