A boy has 3 library cards and 8 books of his interest in the library. Of these 8, he does not want to borrow Chemistry part II unless Chemistry part I is also borrowed. In how many ways can he choose the three books to be borrowed?

Options

A. 56
B. 27
C. 26
D. 41

Answer

To solve the problem of how many ways the boy can choose 3 books from 8, given the condition regarding Chemistry books, we can break it down into cases.

### Step 1: Identify the books and conditions
The boy has 8 books:
- Chemistry Part I (C1)
- Chemistry Part II (C2)
- 6 other books (let's call them B1, B2, B3, B4, B5, B6)

The condition is that he will not borrow Chemistry Part II (C2) unless he also borrows Chemistry Part I (C1).

### Step 2: Create cases based on the condition
We can create two cases based on whether Chemistry Part I is included or not.

**Case 1:** The boy borrows both Chemistry Part I (C1) and Chemistry Part II (C2).
- In this case, he has already chosen 2 books (C1 and C2), and he needs to choose 1 more book from the remaining 6 books (B1, B2, B3, B4, B5, B6).
- The number of ways to choose 1 book from 6 is given by the combination formula \( C(n, r) \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.

\[ \text{Ways in Case 1} = C(6, 1) = 6 \]

**Case 2:** The boy does not borrow Chemistry Part I (C1).
- In this case, he cannot borrow Chemistry Part II (C2) either, so he must choose all 3 books from the remaining 6 books (B1, B2, B3, B4, B5, B6).
- The number of ways to choose 3 books from 6 is:

\[ \text{Ways in Case 2} = C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 3: Calculate the total number of ways
Now, we can add the number of ways from both cases to find the total number of ways the boy can choose 3 books.

\[ \text{Total Ways} = \text{Ways in Case 1} + \text{Ways in Case 2} = 6 + 20 = 26 \]

### Final Answer
The total number of ways the boy can choose 3 books to borrow is **26**.

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