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. if atleast one book on each subjects is taken ?

Options

A. a. 2365
B. b. 3255
C. c. 4224
D. d. 1236

Answer

To solve the problem of how many different collections of books can be made such that at least one book from each subject is included, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Number of Books in Each Subject**:
- Mathematics: 3 books (let's denote them as M1, M2, M3)
- Science: 4 books (denote as S1, S2, S3, S4)
- Literature: 5 books (denote as L1, L2, L3, L4, L5)

2. **Calculate the Total Ways to Select Books from Each Subject**:
- For Mathematics, each book can either be included or not included. Therefore, the total number of ways to select books from Mathematics is \(2^3\) (since there are 3 books).
- For Science, the total number of ways to select books is \(2^4\) (since there are 4 books).
- For Literature, the total number of ways to select books is \(2^5\) (since there are 5 books).

Thus, we have:
\[ \text{Ways to choose from Mathematics} = 2^3 = 8 \]
\[ \text{Ways to choose from Science} = 2^4 = 16 \]
\[ \text{Ways to choose from Literature} = 2^5 = 32 \]

3. **Adjust for the Condition of Selecting at Least One Book**:
- Since we need to ensure that at least one book from each subject is selected, we need to subtract the cases where no books are selected from each subject.
- The case where no books are selected from Mathematics is 1 (choosing none of M1, M2, M3).
- The case where no books are selected from Science is 1 (choosing none of S1, S2, S3, S4).
- The case where no books are selected from Literature is 1 (choosing none of L1, L2, L3, L4, L5).

Therefore, we adjust the calculations:
\[ \text{Valid ways to choose from Mathematics} = 2^3 - 1 = 8 - 1 = 7 \]
\[ \text{Valid ways to choose from Science} = 2^4 - 1 = 16 - 1 = 15 \]
\[ \text{Valid ways to choose from Literature} = 2^5 - 1 = 32 - 1 = 31 \]

4. **Calculate the Total Number of Collections**:
- Now, we multiply the valid selections from each subject to get the total number of collections:
\[ \text{Total collections} = (2^3 - 1) \times (2^4 - 1) \times (2^5 - 1) = 7 \times 15 \times 31 \]

5. **Perform the Multiplication**:
- First, calculate \(7 \times 15 = 105\).
- Then, calculate \(105 \times 31 = 3255\).

Thus, the total number of different collections that can be made such that at least one book from each subject is taken is **3255**.