Question

There are 3 books of mathematics, 4 of science and 5 of literature. How many different collections can be made such that each collection consists of:
1. one book of each subject
2. at least one book of each subject
3. at least one book of literature.

Answer 1

Hint: We here have been given 3 books of mathematics, 4 of science and 5 of literature. For the first part of the question, we will find the number of ways of selecting one book of each subject using the formula for selecting ‘r’ items from ‘n’ items given by $^{n}{{C}_{r}}$ which is expanded as $\dfrac{n!}{r!\left( n-r \right)!}$. Then we will find the product for all the subjects as the selection is to be done simultaneously. Hence, we will get the answer to the first part. Now, for the second part of the question, we will see that there are 2 ways of selecting any book- either it will be selected or not. We will calculate this or all the subjects and then omit from them the cases in which none of the book is selected. Then we will find the product of the resultants for all the subjects. Thus, we will get the answer to the second part. We will use the same for literature books in the third part and we will not omit any cases for mathematics and science as the number of those books doesn’t matter. Then we will find the product of the resultants again. Hence, we will get the required answer of the third part.

Complete step by step answer:
We here have been given 3 books of mathematics, 4 of science and 5 of literature.
In the first part of the question, we have to find out the number of ways in which a collection can be formed such that one book of each subject is included. Hence, we have to select one book from each subject.
Now, we know that the number of ways of selecting ‘r’ items from ‘n’ items is given by $^{n}{{C}_{r}}$ which is expanded as $\dfrac{n!}{r!\left( n-r \right)!}$.
Thus, the number of ways of selecting one book out of 3 books of mathematics is given as:
$^{3}{{C}_{1}}$
Expanding this, we get:
$\begin{align}
  & \dfrac{3!}{1!\left( 3-1 \right)!} \\
 & \Rightarrow \dfrac{3!}{1!2!} \\
 & \Rightarrow \dfrac{3\times 2!}{1!2!} \\
 & \Rightarrow 3 \\
\end{align}$
Thus, there are a total of 3 ways to select 1 book out of 3 books of mathematics.
Now, the number of ways of selecting one book out of 4 book of science is given as:
$^{4}{{C}_{1}}$
Expanding this, we get:
$\begin{align}
  & \dfrac{4!}{1!\left( 4-1 \right)!} \\
 & \Rightarrow \dfrac{4!}{1!3!} \\
 & \Rightarrow \dfrac{4\times 3!}{1!3!} \\
 & \Rightarrow 4 \\
\end{align}$
Thus, there are a total of 4 ways to select 1 book out of 4 books of science.
Now, the number of ways of selecting one book out of 5 book of literature is given as:
$^{5}{{C}_{1}}$
Expanding this, we get:
$\begin{align}
  & \dfrac{5!}{1!\left( 5-1 \right)!} \\
 & \Rightarrow \dfrac{5!}{1!4!} \\
 & \Rightarrow \dfrac{5\times 4!}{1!4!} \\
 & \Rightarrow 5 \\
\end{align}$
Thus, there are a total of 5 ways to select 1 book out of 5 books of literature.
Thus, there are 3, 4 and 5 ways to select one book of mathematics, science and literature respectively. Now, we have to make a collection of them hence the selection of all the subject books have to be done simultaneously and we know that when multiple tasks are to be done simultaneously, the number of ways to do that is equal to the product of the number of ways of doing those tasks separately.
Thus, the number of ways of making the required collection is given as:
$\begin{align}
  & 3\times 4\times 5 \\
 & \therefore 60 \\
\end{align}$
Thus, the number of ways of making a collection which consists of one book of each subject is 60.
Now, in the second part of the question, we have to find the number of collections which consists of at least one book of all the subjects.
Now, we know that for a book to be a part of the collection, there will always be 2 cases. Either the book will be a part of the collection or it won’t be. Thus, there will come out to be 2 cases for every book.
Now, let us first take the case of 3 books of mathematics. Here, each of the three books can either be selected or it can’t be and the selection of each book is independent of one another. But among the total cases, there will also come out a case in which none of the book is selected and that case is to be removed from the total number of cases.
Thus, the number of ways of selecting at least 1 book out of 3 books of mathematics is given as:
$\begin{align}
  & \left( 2\times 2\times 2 \right)-1 \\
 & \Rightarrow {{2}^{3}}-1 \\
\end{align}$
If we use the similar formula for 4 for books of science, we will get the following number of ways to select one book out of 4 books is given as ${{2}^{4}}-1$ and one book out of 5 books of literature is ${{2}^{5}}-1$.
Now, for a collection these books have to be selected simultaneously, thus, the number of ways of making this collection is:
$\begin{align}
  & \left( {{2}^{3}}-1 \right)\left( {{2}^{4}}-1 \right)\left( {{2}^{5}}-1 \right) \\
 & \Rightarrow \left( 8-1 \right)\left( 16-1 \right)\left( 32-1 \right) \\
 & \Rightarrow 7\times 15\times \times 31 \\
 & \therefore 3255 \\
\end{align}$
Thus, the number of collections such that at least one book of all the subjects is in there is 3255.
Now, for the third part of this question, we need to find the number of collections which consist of at least one literature book. Thus, it doesn’t matter how many mathematics or science books are there but there should be at least one book of literature.
Now, as mentioned above, the number of ways of selecting at least 1 book of literature is given as:
${{2}^{5}}-1$
Now, if we take the case of mathematics and science books, it doesn’t matter how many there are they. They can be 0 for all that matters. Thus, the number of ways of selecting mathematics books is ${{2}^{3}}$ and science books is ${{2}^{4}}$.
Now, these selections are to be done simultaneously. Hence, the total number of these collections is given as:
$\begin{align}
  & {{2}^{3}}\times {{2}^{4}}\times \left( {{2}^{5}}-1 \right) \\
 & \therefore 3968 \\
\end{align}$

Thus, the total number of collections such that they contain at least one book of literature is 3968.

Note: Here in the first part, we have calculated the number of ways of selecting one book from each subject separately first and then multiplied them. But it may be more time consuming for a similar case with more number of objects. So we may calculate them altogether by keeping them in the symbolical form until the final calculation and then expand all of them together.