JEE Main 2026 — 8 April, Evening Shift — Question 27

Method 1: Principle of Inclusion-Exclusion (PIE) This method involves calculating the total number of distributions and subtracting the cases where one or more bags are empty.

Total Distributions: Each of the 4 distinct books can be placed into any of the 3 distinct bags.

$$\text{Total} = 3^4 = 81$$

Cases with at least one empty bag: We choose 1 bag to be empty in $\binom{3}{1}$ ways and distribute the 4 books into the remaining 2 bags in $2^4$ ways.

$$\text{Subtotal}_1 = \binom{3}{1} \times 2^4 = 3 \times 16 = 48$$

Cases with at least two empty bags: We choose 2 bags to be empty in $\binom{3}{2}$ ways and distribute the 4 books into the remaining 1 bag in $1^4$ ways.

$$\text{Subtotal}_2 = \binom{3}{2} \times 1^4 = 3 \times 1 = 3$$

Applying PIE Formula:

$$\text{Required Ways} = \text{Total} - (\text{At least 1 empty}) + (\text{At least 2 empty})$$ $$\text{Required Ways} = 81 - 48 + 3 = 36$$

Method 2: Grouping and Partitioning We can partition the 4 distinct books into 3 non-empty groups and then arrange those groups among the 3 bags. The only possible distribution size for 4 items into 3 non-empty groups is $(2, 1, 1)$.

Grouping the books: The number of ways to divide 4 distinct objects into groups of size 2, 1, and 1 is:

$$\text{Ways to Group} = \frac{4!}{2!1!1! \times 2!} = \frac{24}{2 \times 2} = 6$$

(Note: We divide by $2!$ because two groups are of equal size 1.)

Distributing to bags: Since the bags are distinct, we arrange these 3 groups in the 3 bags in $3!$ ways.

$$\text{Total Ways} = 6 \times 3! = 6 \times 6 = 36$$