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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

Updated On: Jan 15, 2026
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Correct Answer: 1000

Solution and Explanation

Distribution Problem: Balloons, Pencils, and Erasers

Given:
- 15 identical balloons.  
- 6 identical pencils. 
- 3 identical erasers. 
- 3 children.

Each child requires a minimum of 4 balloons and 1 pencil. We will first allocate these minimums.

Balloon Distribution:

Each of the 3 children receives 4 balloons, totaling \(3 \times 4 = 12\) balloons. This leaves \(15 - 12 = 3\) balloons remaining for further distribution.

To find the number of ways to distribute \(n\) identical objects among \(r\) recipients, we use the stars and bars formula: \(\binom{n+r-1}{r-1}\).

For the remaining 3 balloons (\(n=3\)) and 3 children (\(r=3\)):

Number of ways to distribute remaining balloons: \(\binom{3+3-1}{3-1} = \binom{5}{2} = \frac{5!}{2!3!} = 10\)

Pencil Distribution:

Each of the 3 children receives 1 pencil, totaling \(3 \times 1 = 3\) pencils. This leaves \(6 - 3 = 3\) pencils remaining.

Using the same formula for the remaining 3 pencils (\(n=3\)) among 3 children (\(r=3\)):

Number of ways to distribute remaining pencils: \(\binom{5}{2} = 10\)

Eraser Distribution:

There are 3 identical erasers (\(n=3\)) to be distributed among 3 children (\(r=3\)).

Number of ways to distribute erasers: \(\binom{5}{2} = 10\)

Total Distribution Ways:

The total number of ways to distribute all items is the product of the ways for each item:

\(Total = 10 \, (\text{balloons}) \times 10 \, (\text{pencils}) \times 10 \, (\text{erasers})\) 
\(Total = 1000\)

Therefore, there are 1000 distinct ways to distribute the items to the children according to the specified conditions.

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