Question

Four different mathematics books, six different physics books and two different chemistry books are to be arranged on a shelf. How many different arrangements are possible if -
A) The book in a particular subject must all stand together
B) Only the mathematics books must stand together?

Answer 1

In this question, we are given four mathematics books, six physics books and two chemistry books, and we are asked to arrange them in a certain way. We will use factorials for this. In part (a), first group the similar kind of books together and then arrange them. Then, arrange the books within the groups to find the answer. In part (b), group only mathematics books and then arrange all the books. Then, arrange the mathematics books within themselves.

Complete step-by-step answer:
We are given four different mathematics books, six different physics books and two different chemistry books.
A) There are two parts to this question. In the first part, we will look at the subject books as a whole. Since we are asked to keep the books in a particular subject together, we will assume all the mathematics books to be one single element, all the physics books as one element and all the chemistry books as one element. Now, we have three different elements which can be arranged in $3!$ ways.
Now, within each single element of these books, there are the actual number of books. For example, in one element in mathematics, there are four books. These books will arrange within themselves in $4!$ ways.
Similarly, in one element of physics, there are six books. The books will arrange within themselves in $6!$ ways.
In one element of chemistry, there are two books which will arrange within themselves in $2!$ ways.
Therefore, total possible arrangements= $3! \times 4! \times 6! \times 2!$ = $6 \times 24 \times 720 \times 2 = 2,07,360$ways.
B) In this part, we are only required to keep all the math books together. We will follow the same approach as above and consider all the mathematics books as one single element. Now, there are 6 physics books, 2 chemistry books and 1 element of math books. In total, we have $6 + 2 + 1 = 9$ books now.
These 9 books can be arranged in $9!$ ways. After this, we will look at that one element of math books. These books will arrange themselves in $4!$ ways because there are 4 books.
Now, total arrangements = $9! \times 4! = 3,62,880 \times 24 = 87,09,120$

Note: Notice the word ‘different’ given in the question. If this word had not been there, all the books would have been considered similar and there would not have been any requirement to arrange those. For example: in part (a), the answer would have been $3!$ as the books within each element would have been the same.