Question

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

Answer 1

Given:
- 15 identical balloons.
- 6 identical pencils.
- 3 identical erasers.
- 3 children.
Each child must get at least 4 balloons and 1 pencil. First, let's distribute the minimum required balloons and pencils to each child:
For the Balloons:
Each child gets 4 balloons. So, for 3 children: $3 \times 4 = 12$ balloons are given. We're left with $15 - 12 = 3$ balloons to be distributed.
Now, let's use the formula for distributing $n$ identical objects among $r$ people/groups.
The formula is:$(^{n+r-1}C_{r-1})$
Where: - $n$ = number of identical objects - $r$= number of groups/people
Here, $n = 3$ (remaining balloons) and $r = 3$ (children).

Number of ways to distribute 3 identical balloons among 3 children
= $^{3+3-1}C_{3-1}$
$= \space^{5}C_{2}$
$= \frac{5!}{2!3!}$
$= 10$
For the Pencils: Each child gets 1 pencil.
So, for 3 children: $3 \times 1 = 3$ pencils are given.
We're left with $6 - 3 = 3$ pencils to be distributed.
Using the formula again, for $n = 3$ pencils among $r = 3$ children:
Number of ways to distribute 3 identical pencils among 3 children = $^{5}C_{2} = 10$

For the Erasers: There are 3 identical erasers and 3 children.
So, using the formula for $n = 3$ erasers and $r = 3$ children:
Number of ways to distribute 3 identical erasers among 3 children = $^{5}C_{2} = 10$

Now, the total number of ways is the product of all the individual ways:
$Total = 10 (for \space balloons) \times 10 (for\space pencils) \times 10 (for \space erasers)$
$Total = 1000$

So, there are 1000 ways to distribute the items among the children satisfying the given conditions.