CAT Exam 2021 Slot 2 Solved Paper

This question is an application of the product rule in probability and combinatorics.
In the product rule, if two events A and B can occur in x and y ways, and for an event E, both events A and B need to take place, the number of ways that E can occur is $\mathit{x}\mathit{y}$. This can be expanded to 3 or more events as well.
Event 1: Distribution of balloons
Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them.
So we are left with $15-4\times 3=15-12=3\mathrm{balloons}\mathrm{and}3\mathrm{children}$.
Now we need to distribute 3 identical balloons to 3 children.
So we are left with $15-4\times 3=15-12=3\mathrm{balloons}\mathrm{and}3\mathrm{children}$
Now we need to distribute 3 identical balloons to 3 children.
This can be done in ${}_{}^{\mathrm{n}+\mathrm{r}-1}{\mathrm{C}}_{\mathrm{r}}^{}-1$ ways, where $\mathit{n}=3\mathrm{and}\mathit{r}=3$
so, number of the ways - ${}_{}^{3+3-1}{\mathrm{C}}_3^{}-1{=}^5{\mathrm{C}}_2=\frac{5\times 4}{2\times 1}=10$
Event 2: Distribution of pencils
Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with $6-3=3\mathrm{pencils}$.
We now need to distribute 3 identical pencils to 3 children. This can be done in ${}_{}^{\mathrm{n}+\mathrm{r}-1}{\mathrm{C}}_{\mathrm{r}}^{}-1$ ways, where $\mathit{n}=3\mathrm{and}\mathit{r}=3$
so, number of the ways = ${}_{}^{3+3-1}{\mathrm{C}}_3^{}-1{=}^5{\mathrm{C}}_2=\frac{5\times 4}{2\times 1}=10$
Event 3: Distribution of erasers
We need to distribute 3 identical erasers to 3 children.
This can be done in ${}_{}^{\mathrm{n}+\mathrm{r}-1}{\mathrm{C}}_{\mathrm{r}}^{}-1$ ways, where n = 3 and \r = 3
so, number of the ways = ${}_{}^{3+3-1}{\mathrm{C}}_3^{}-1{=}^5{\mathrm{C}}_2=\frac{5\times 4}{2\times 1}=10$
Applying the product rule, we get the total number of ways $=10\times 10\times 10=1000$.