This problem is an application of the non negative integer solution and the symmetry argument. This is from ISI MStat 2015 PSB Problem 4.

## Problem

Suppose 15 identical balls are placed in 3 boxes labeled A, B and C. What is the number of ways in which Box A can have more balls than

Box C?

### Prerequisites

- Symmetry Argument
- Number of non negative integer solutions to \(x+y+z = 15\) is \(17 \choose 2\). Not sure why? Search it up!

## Solution ( No Algebra )

There are three possible cases.

- Box A has more balls than Box C.
- Box C has more balls than Box A.
- Box A has the same number of balls as Box C.

**The symmetry argument**

The number of ways in which Box A has more balls than Box C = The number of ways in which Box C has more balls than Box A.

Isn’t that very obvious, since the balls are not biased towards boxes. Why will they be?

Total Number of Ways = The number of ways in which Box A has more balls than Box C + The number of ways in which Box C has more balls than Box A + The number of ways in which Box A has the same number of balls as Box C.

The number of ways in which Box A has the same number of balls as Box C = 8 right?

Why? They can have either 0 ball each, 1 ball each, 2 balls each, …, 7 balls each at most.

Therefore, The number of ways in which Box A has more balls than Box C = \(\frac{{17 \choose 2 }- 8}{2} = 64 = \frac{(n+1)^2}{2}\). [\( n = 15\)]

### Challenge Problem

Suppose 15 identical balls are placed in 3 boxes labeled A, B, and C.

What is the number of ways in which Box A have no fewer balls than

Box B and Box B have no fewer balls than Box C?

This is related to the topic of mathematics called **Partitions of Numbers**.

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