Question

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is:

• A. \(18 \)
• B. \(36 \)
• C. \(39 \)
• D. \(72 \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem asks for the number of ways to distribute 4 distinct objects (books) into 3 distinct containers (bags) such that no container is left empty. This is equivalent to finding the number of onto (surjective) functions from a set of size 4 to a set of size 3.
Step 2: Key Formula or Approach:
There are two primary methods:
Method 1: Using the Principle of Inclusion-Exclusion formula:
Total onto functions = \( \sum_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m \), where \( m \) is the number of items and \( n \) is the number of boxes.
Method 2: Grouping and distribution. Divide the 4 books into 3 groups and then arrange them in the 3 bags.
Step 3: Detailed Explanation:
Let's use the grouping and distribution method.
Since we have 4 distinct books and 3 distinct bags, and no bag can be empty, the number of books in the bags must follow the distribution pattern 2, 1, 1.
First, divide the 4 distinct books into 3 groups of sizes 2, 1, and 1.
The number of ways to divide \( n \) distinct items into groups of size \( p, q, r \) is \( \frac{n!}{p! q! r! \cdot k!} \), where \( k \) is the number of groups of identical sizes.
Here, the sizes are 2, 1, 1. We have two groups of size 1, so \( k = 2 \).
Number of ways to form the groups = \( \frac{4!}{2! 1! 1! 2!} = \frac{24}{2 \cdot 1 \cdot 1 \cdot 2} = \frac{24}{4} = 6 \).
Now we have 3 distinct groups of books. We need to distribute them into 3 distinct bags.
Number of ways to arrange 3 groups into 3 bags is \( 3! = 6 \).
Total number of ways = (Number of ways to group) \( \times \) (Number of ways to distribute)
Total ways = \( 6 \times 6 = 36 \).
Alternatively, using Inclusion-Exclusion:
Total ways without restriction = \( 3^4 = 81 \).
Subtract ways where at least 1 bag is empty = \( \binom{3}{1} \times 2^4 = 3 \times 16 = 48 \).
Add ways where at least 2 bags are empty (since they were subtracted twice) = \( \binom{3}{2} \times 1^4 = 3 \times 1 = 3 \).
Number of ways = \( 81 - 48 + 3 = 36 \).
Step 4: Final Answer:
The total number of ways is 36.