CAT 2021

Slot 2

QA

Question & Solution

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 xy. 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 x 3 = 15 - 12 = 3 balloons and 3 children. 

Now we need to distribute 3 identical balloons to 3 children. 

This can be done in $^{n+r-1}C_{r-1}$ ways, where n = 3 and r = 3. 

So, number of ways = $^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$

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 pencils.

We now need to distribute 3 identical pencils to 3 children.

This can be done in $^{n+r-1}C_{r-1}$ ways, where n = 3 and r = 3.

So, number of ways = $^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$

Event 3: Distribution of erasers

We need to distribute 3 identical erasers to 3 children.

This can be done in $^{n+r-1}C_{r-1}$ ways, where n = 3 and r = 3.

So, number of ways = $^{3+3-1}C_{3-1}=^5C_2=\frac{5\times\ 4}{2\times\ 1}=10$

Applying the product rule, we get the total number of ways = 10 x 10 x 10 = 1000.