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

Answer

To solve the problem, we need to consider the conditions given regarding the borrowing of the books. The boy has 8 books, but he has specific conditions for borrowing Mathematics Part I (M1) and Mathematics Part II (M2). Let's denote the books as follows:

- M1: Mathematics Part I
- M2: Mathematics Part II
- B1, B2, B3, B4, B5, B6: Other 6 books

The boy has 3 library tickets, and we need to find out how many ways he can choose 3 books to borrow under the given conditions.

### Step 1: Consider the cases based on the borrowing of M1 and M2.

**Case 1: The boy borrows M1.**
- If he borrows M1, he must also borrow M2.
- So, he has already chosen 2 books (M1 and M2), 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:
\[ \text{Ways} = \binom{6}{1} = 6 \]

**Case 2: The boy does not borrow M1.**
- If he does not borrow M1, he cannot borrow M2 either.
- In this case, he needs to 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 given by:
\[ \text{Ways} = \binom{6}{3} = \frac{6!}{3! \cdot (6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 2: Add the number of ways from both cases.

Now, we add the number of ways from both cases to get the total number of ways the boy can choose the books:
\[ \text{Total Ways} = \text{Ways from Case 1} + \text{Ways from Case 2} = 6 + 20 = 26 \]

### Final Answer:
The boy can choose the three books to be borrowed in **26 different ways**.

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