### Instruction: At the grocery store, I bought $4$ different items. I brought $3$ identical bags, and handed them to the cashier. How many ways are there for the cashier to put the items I bought in the $3$ identical bags, assuming he might leave some of the bags empty?### Response: This problem is about counting the number of partitions of a set of $4$ elements into $3$ subsets, where the order of the subsets does not matter.
One way to approach this is to use the stars and bars method, where we represent the items as stars and the bag boundaries as bars.
For example, one possible partition is **|*|*, which means two items in the first bag, one item in the second bag, and one item in the third bag.
Another possible partition is ****||, which means all four items in the first bag, and the other two bags empty.
To count the number of partitions, we need to arrange $4$ stars and $2$ bars in a row, where the order matters.
There are $6$ positions to choose from, and we need to choose $2$ of them for the bars.
The number of ways to do this is $\binom{6}{2} = \frac{6!}{2!4!}
= 15$.
Therefore, there are $15$ ways for the cashier to put the items in the bags.