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:
This is a problem of distributing distinct objects into distinct boxes such that no box is empty. This can be solved using the principle of Inclusion-Exclusion or by partitioning the books.
Step 2: Key Formula or Approach:
1. Total ways to put $n$ distinct books in $r$ distinct bags = \(r^n\).
2. Number of onto functions (surjective) from $n$ elements to $r$ elements = \(r^n - \binom{r}{1}(r-1)^n + \binom{r}{2}(r-2)^n \dots\)
Step 3: Detailed Explanation:
Here, $n=4$ (books) and $r=3$ (bags). Total ways without restriction = \(3^4 = 81\). Ways where at least one bag is empty: - Choose 1 bag to be empty: \(\binom{3}{1} \times 2^4 = 3 \times 16 = 48\). - Choose 2 bags to be empty: \(\binom{3}{2} \times 1^4 = 3 \times 1 = 3\). By Inclusion-Exclusion: \[ \text{Ways} = 81 - [48 - 3] = 81 - 45 = 36 \] Alternatively, partition 4 books into 3 groups: The only way to partition 4 into 3 non-empty sets is $\{2, 1, 1\}$. Number of ways = \(\frac{4!}{2!1!1!2!} \times 3! = 6 \times 6 = 36\).
Step 4: Final Answer:
The number of ways is 36.