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.

Answer 2

There are 15 identical balloons, 6 identical pencils, and 3 identical erasers.
We need to distribute these items among 3 children, such that each child gets at least 4 balloons and 1 pencil.
Therefore, every person gets 4 balloons and 1 pencil.
After distribution, we have 3 balloons, 3 pencils, and 3 erasers remaining.
By arranging the balloons in all possible combinations.
(3, 0, 0) , (2, 1, 0) , (1, 1, 1)
We can arrange (3, 0, 0) in three ways.
We can arrange (2, 1, 0) in six ways.
We can arrange (1, 1, 1) in only one way.
So, we can arrange balloons among three children in 10 ways.
Likewise, we can also arrange 3 pencils in 10 ways and 3 erasers in 10 ways.

Therefore, the total number of ways is 10 x 10 x 10 = 1000 ways