Question

There are 6 books of economics, 3 of mathematics and 2 of accountancy. In how many ways can these be placed on a shelf, if:
1. Books on the same subject are together?
2. Books on the same subject are not together?

Answer 1

Hint: We here have been given 6 books of economics, 3 of mathematics and 2 of accountancy and we have to arrange them in such a way in the first part such that the books of the same subjects are together. For this, we will first form a separate unit of books of different subjects. Then we will use the formula to arrange ‘n’ distinct objects given as n!. Then, inside the units themselves, we will also arrange those books simultaneously with the units of books and hence we will get the answer of the first part. For the second part, we have to calculate the number of ways to arrange these books such that books of the same subjects are not together. This is the complete opposite of the first part. So for this, we will find the number of ways in which all these books can be arranged and then we will subtract from it the number of ways in which the books of the same subjects are together. Thus, we will get the answer to the second part.

Complete step by step answer:
Here, we have been given 6 books of economics, 3 of mathematics and 2 of accountancy. In the first part of the question, we have to arrange the books in such a way that the books of the same subject are together.
Thus, for this, we will form separate units of books of different subjects. Now, if we assume the 6 books of economics to be \[{{E}_{1}},{{E}_{2}},{{E}_{3}},...,{{E}_{6}}\], the 3 books of mathematics to be ${{M}_{1}},{{M}_{2}},{{M}_{3}}$ and the 2 books of accountancy to be ${{A}_{1}},{{A}_{2}}$, then the there will be 3 different units of books formed as we have three different subjects.
Now, we know that the number of ways to arrange ‘n’ distinct objects is given by n!.
Thus, the number of ways to arrange these 3 units of books is given as:
$3!$
Now, in these 3 units, the books itself are distinct. Thus, there will be more than one way to arrange these books among their own units.
Now, we know that there are 6 books of economics, thus the number of ways to arrange the unit of economics books is:
$6!$
Now, we also know that there are 3 books of mathematics, thus the number of ways to arrange the unit of mathematics books is:
$\begin{align}
  & 3! \\
 & \\
\end{align}$
Similarly, we also know that there are 2 books of accountancy, thus the number of ways to arrange the unit of accountancy books is:
$2!$
Now, we know that all of these arrangements are to be done simultaneously and when different arrangements are to be done simultaneously, we multiply the number of ways those arrangements can be done separately. Thus, the required number of ways are given as:
$3!\times 6!\times 3!\times 2!$
Expanding these factorials, we get:
$\begin{align}
  & 6\times 720\times 6\times 2 \\
 & \therefore 51840 \\
\end{align}$
Thus, the required number of ways to arrange these books such that the books of the same subjects are together is 51840.
Now, we have to find the number of ways to arrange these books in such a way that the books of the same subjects are not together.
We know that this can be found out by subtracting from the total number of ways to arrange these books the number of ways to arrange these books such that the books of the same subjects are together.
Now, we have already found out one required thing, i.e. the number of ways to arrange these books such that the books of the same subjects are together in the first part. Now, we will find the total number of ways to arrange these books.
We have here 6 books of economics, 3 of mathematics and 2 of accountancy. Thus the total books are given as:
$6+3+2=11$
Thus, there are a total of 11 books and all of them are different. Thus, the total number of ways to arrange these books is given as:
$\begin{align}
  & 11! \\
 & \Rightarrow 39916800 \\
\end{align}$
Thus, there are a total of 39916800 ways to arrange these books.
Now, the total number of ways to arrange these books such that the books of the same subjects are not together are given as:
$\begin{align}
  & 3996800-51840 \\
 & \therefore 39864960 \\
\end{align}$

Thus, there are a total of 39864960 ways to arrange these books such that the books of the same subject are not together.

Note: We here have calculated the values of the factorials in the answer with the help of a calculator. But practically, these calculations are long and tedious so while in exam, leave the answer in the factorial form only as there are no calculators available to us during the exam so the answer in factorial form would be enough.