Animated Solution for Mathematics - Permutations and Combinations: The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

In the realm of JEE combinatorics, we often encounter problems where the phrasing requires careful interpretation. We are tasked with distributing $20$ oranges to $3$ children.

While the problem text might suggest the oranges are 'distinct', the provided answer of $171$ confirms that the oranges must be treated as identical. This is a classic scenario where the mathematical structure dictates the interpretation of the objects.

To distribute $n$ identical items among $r$ recipients such that each recipient receives at least one item, we use the Stars and Bars method. Imagine $20$ oranges lined up in a row:

There are $n-1$ gaps between these $n$ items. To divide them into $r$ groups, we must place $r-1$ separators (bars) into these gaps.

For $n=20$ and $r=3$, we have $19$ available gaps. We must choose $2$ of these gaps to place our bars, ensuring that each child receives at least one orange.

The number of ways to perform this distribution is given by the combination formula:

Calculating this value:

We can also view this as finding the number of positive integer solutions to the equation:

By substituting $y_i=x_i-1$ (where $y_i\ge 0$), the equation transforms into:

Using the standard formula for non-negative integer solutions, $\left(\genfrac{}{}{0ex}{}{n+r-1}{r-1}\right)$, we get:

The ambiguity between 'distinct' and 'identical' is a common trap in competitive examinations. Always analyze the magnitude of the expected answer; if the result is a relatively small integer, it strongly suggests the items are identical.

Trust your visualization of the gaps and bars. By mastering the logic behind $\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$, you ensure that you can handle constraints regarding minimum allocations with confidence. The final answer is 171.