Question

A person has 3 different bags & 4 different books. The number of ways in which he can put these books in the bags so that no bag is empty, is

• A. 36
• B. 24
• C. 32
• D. 30

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to distribute 4 distinct books into 3 distinct bags such that no bag is empty.
Since each bag must have at least 1 book and there are 4 books total, the only possible distribution pattern (partition of 4 into 3 positive parts) is $(1, 1, 2)$, meaning one bag gets 2 books and the other two bags get 1 book each.
Step 2: Counting the Arrangements:
Step 2a: Choose which bag gets 2 books:
There are $\binom{3}{1} = 3$ ways to choose which of the 3 bags receives 2 books.
Step 2b: Choose 2 books out of 4 for that bag:
There are $\binom{4}{2} = 6$ ways to choose 2 books from the 4 for the selected bag.
Step 2c: Distribute remaining 2 books to the remaining 2 bags:
The remaining 2 books go to 2 distinct bags, one book each.
There are $2! = 2$ ways to assign these 2 books to the 2 remaining bags.
Step 3: Total Count:
\[ \text{Total} = \binom{3}{1}\times\binom{4}{2}\times 2! = 3 \times 6 \times 2 = 36 \] Alternatively, using the multinomial directly: \[ \frac{4!}{2!\cdot 1!\cdot 1!} \times 3 = \frac{24}{2} \times 3 = 12 \times 3 = 36 \] Step 4: Final Answer:
The number of ways is $\mathbf{36}$.
The answer is Option (1).