JEE Main 2026 Apr 8 Shift 2, Mathematics Q9: Combinations

With $4$ distinct books into $3$ distinct bags and no bag empty, the only book-count distribution per bag is $(2,1,1)$ in some order.

Step 1: Choose which $2$ books go together: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways.

Step 2: The remaining $2$ books go individually into the other two bags: $2!$ orderings.

Step 3: Assign the three groups (one pair, two singletons) to the three labelled bags: $3!$ ways. But we have already differentiated the two singletons in step $2$, so combine carefully — use the cleaner form:

$\text{Total}=\frac{4!}{2!1!1!}\cdot \frac{3!}{2!}\cdot \frac{1}{1}$

$=12\cdot 3=36$.

(Here $4!/(2!1!1!)=12$ distributes labelled books into three labelled groups of sizes $2,1,1$, and $3!/2!=3$ accounts for the choice of which bag gets the pair.)

$\therefore$ $36$ ways.