Question

A boy has \[3\] library tickets and \[8\] books of his interest in the library. Out of these \[8\], he does not want to borrow Chemistry part II, unless Chemistry part I is also borrowed. The number of ways in which he can choose the three books to be borrowed is:

Answer 1

Hint:As per the given information, the boy could borrow the book in two ways. We will find the probabilities of these two different cases. Then we will add these two probabilities and get the final probability.

Complete step-by-step answer:
It is given that; a boy has \[3\] library tickets and \[8\] books of his interest in the library. Out of these \[8\], he does not want to borrow Chemistry part II, unless Chemistry part I is also borrowed.
We have to find the number of ways in which the boy can choose the three books to be borrowed.
There are two different possibilities in which the boy can borrow three books. These are:
When he selected the Chemistry part II book, he also borrowed the Chemistry part I book, and the third book is selected from the remaining \[6\] books.
When the Chemistry part II is not selected, in this case he selected the three books from the remaining \[7\] books.
For, the first case, the books can be borrowed in following way: \[^6{C_1} = 6\] ways
For second case, the books can be borrowed in following way: \[^7{C_3} = \dfrac{{7 \times 6 \times 5}}{{1 \times 2 \times 3}} = 35\] ways
So, the total number of ways in which he can borrow the books is \[35 + 6 = 41\] ways.
$\therefore $The number of ways in which he can choose the three books to be borrowed is: \[41\] ways.

Note:We know that the number of ways \[r\] number of objects can be drawn from \[n\] objects is \[^n{C_r}\].
\[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
Here, we can consider two different cases to calculate the probabilities.