Question

The school library has 50 action adventure novels, 15 romances and 10 historical novels. Julie wants to take one of each type for her sick cousin to read. How many different choices of three books are available to her?

Answer 1

Hint: In this question, first of all choose 1 novel from each category and then multiply them to get the total numbers of different choices of three books available to her. We will be using the formula for selecting the books as \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.\] We will substitute n=50 and r=1 for selecting 1 book from 50 action adventure novels. Similarly, we will do it for romance and historical novels. Then, to get the final answer, we will multiply the obtained results.

Complete step-by-step answer:
In this question, we are given a school library that has 50 action adventure novels, 15 romances, and 10 historical novels. Julie wants to take one of each type for her sick cousin to read. So, we have to find the total number of choices of three books available to her.
We are given that Julie wants to take one book of each type. So, let us first find the ways of selecting one book of each type. We know that the number of ways of selecting r out of n things is given by \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.\]
So, we have a total of 50 action adventure novels. So, the number of ways of selection of 1 out of these is given by.
\[N\left( A \right)={{\text{ }}^{50}}{{C}_{1}}=\dfrac{50!}{49!1!}\]
\[\Rightarrow N\left( A \right)=\dfrac{50\times 49!}{49!}\]
\[\Rightarrow N\left( A \right)=50\text{ ways}......\left( i \right)\]
Also, we have a total of 15 romantic novels. So, the number of ways of selecting 1 out of these is given by,
\[N\left( R \right)={{\text{ }}^{15}}{{C}_{1}}=\dfrac{15!}{1!14!}\]
\[\Rightarrow N\left( R \right)=\dfrac{15\times 14!}{14!}\]
\[\Rightarrow N\left( R \right)=15\text{ ways}......\left( ii \right)\]
Also, we have a total of 10 historical novels. So, the number of ways of selecting 1 out of these is given by,
\[N\left( H \right)={{\text{ }}^{10}}{{C}_{1}}=\dfrac{10!}{1!9!}\]
\[\Rightarrow N\left( H \right)=\dfrac{10\times 9!}{9!}\]
\[\Rightarrow N\left( H \right)=10\text{ ways}......\left( iii \right)\]
Now, we know that according to the multiplication principle of combination, the total number of ways of occurrence of independent events is given by the multiplication of their individual number of ways.
Now, we know that choosing an adventure novel, a romance novel and historical novel is not dependent on each other or independent of each other. So, by using the multiplication principle, we get,
\[\text{Total ways of selecting one novel of each type }N\left( T \right)=N\left( A \right).N\left( R \right).N\left( H \right)\]
By substituting the values of N(A), N(R) and N(H) from equations (i), (ii) and (iii) respectively, we get,
\[N\left( T \right)=50\times 15\times 10=7500\text{ ways}\]
Hence, Julie has a total of 7500 choices of three books available to her.

Note: In this question, many students make this mistake of first finding the total number of novels that 50 (adventure novels) + 15 (romance novels) + 10 (historical novels) = 75 novels and then choosing 3 out of 75 by using \[^{75}{{C}_{3}}\] which is wrong because we have to select 1 novel of each type but by using \[^{75}{{C}_{3}},\] we can ensure that each one of the 3 novels is of a different type.