JEE Mathematics 2023 Question with Solution

Given: $20$ oranges are to be given to $3$ children, with each child getting at least one orange.

Find: The number of ways of distribution.

The solution states:

and then concludes the count is

However, that counting corresponds to distributing identical objects, whereas the oranges here are distinct. For distinct oranges, each orange can be assigned to any one of the $3$ children.

Total assignments without restriction:

Subtract assignments where at least one child gets no orange using inclusion-exclusion:

• If one specified child gets none, the oranges can be distributed among the remaining $2$ children in

ways. Since there are $3$ choices of the omitted child, subtract

Add back assignments where two specified children get none. Then all $20$ oranges go to the remaining one child, which gives

way for each choice. There are

such choices, so add

Therefore,

Therefore, the number of ways is $3483638676$.

Given: the solution uses

and counts

This method counts only how many triples

are possible, where $x$, $y$, and $z$ are the numbers of oranges received by the three children.

But once a triple such as

is fixed, the oranges are still distinct, so we must also count how many different sets of actual oranges go to each child. Hence the stars-and-bars count alone is not the final answer.

That is why the correct result must be obtained by counting onto functions from the set of $20$ distinct oranges to the set of $3$ children, which gives