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
15−4×3=15−12=3balloonsand3children.
Now we need to distribute 3 identical balloons to 3 children.
So we are left with
15−4×3=15−12=3balloonsand3children Now we need to distribute 3 identical balloons to 3 children.
This can be done in
n+r−1Cr−1 ways, where
n=3andr=3 so, number of the ways -
3+3−1C3−1=5C2==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=3pencils.
We now need to distribute 3 identical pencils to 3 children. This can be done in
n+r−1Cr−1 ways, where
n=3andr=3 so, number of the ways =
3+3−1C3−1=5C2==10 Event 3: Distribution of erasers We need to distribute 3 identical erasers to 3 children.
This can be done in
n+r−1Cr−1 ways, where n = 3 and \r = 3
so, number of the ways =
3+3−1C3−1=5C2==10 Applying the product rule, we get the total number of ways
=10×10×10=1000.