CAT 2021 Slot 2 - Quant Question 5

Question 5

The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is


Correct Answer: 1000

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.


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