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

Correct Answer: 1000

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.