## Problem

There are $$6$$ identical gifts to be distributed among $$8$$ children. Each child may get more than $$1$$ gift. How many ways are there to distribute the $$6$$ gifts among the children?

## Solution

One approach is to apply the Stars and Bars method, which involves counting the number of ways one can place "bars" to divide stars (or in case, gifts) among children.

In this case, there are $$8-1=7$$ bars, which will divide the stars into $$8$$ buckets (one per child). Additionally, there are $$6+7=13$$ slots where the bars and stars can be placed. Below is an example configuration:

$|\;|\;|\;|\;|**|***|*$

In the above example, the $$1$$st – 5th children get no presents, the 6th gets $$2$$ presents, the 7th gets $$3$$ presents, and the 8th gets $$1$$ present.

In general, we can count the number of configurations of $$n$$ indistinguishable objects divided between $$k$$ buckets as follows:

$Count = \binom{n+k-1}{k-1}$

For this particular problem, this method gives the following solution:

$Count = \binom{6+8-1}{8-1} = \binom{13}{7} = 1716$