Problem:
In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

Solution:
To solve this problem, we can use a combination of techniques such as stars and bars and inclusion-exclusion principle.

Step 1: Distributing one eraser to each kid
Since each kid must receive at least one eraser, we can start by distributing one eraser to each kid. This leaves us with 7 - 4 = 3 erasers to distribute.

Step 2: Distributing the remaining erasers
Now, we need to distribute the remaining 3 erasers among the 4 kids. However, each kid can receive a maximum of 3 erasers.

To simplify the problem, let's consider the case where there are no restrictions on the number of erasers each kid can receive. In this case, we can use the stars and bars technique to distribute the erasers.

Imagine that we have 3 stars (representing the 3 erasers) and 3 bars (representing the 4 kids). We can arrange them in such a way that each arrangement represents a distribution of erasers. For example,

* | * * | * represents the distribution where the first kid receives 0 erasers, the second kid receives 2 erasers, and the third and fourth kids receive 1 eraser each.

Using stars and bars, the number of ways to arrange the 3 stars and 3 bars is given by (3+3) choose 3 = 6 choose 3 = 6! / (3! * 3!) = 20.

Step 3: Applying the restriction
However, we need to apply the restriction that no kid can receive more than 3 erasers. To do this, we can use the inclusion-exclusion principle.

Let's consider the case where one or more kids receive more than 3 erasers.

Case 1: One kid receives all 3 erasers.
There are 4 ways to choose the kid who receives all 3 erasers. For each of these cases, we are left with 3 erasers to distribute among the remaining 3 kids. Using stars and bars, the number of ways to distribute the remaining erasers is 5 choose 2 = 10.

Case 2: Two kids receive all 3 erasers.
There are 4 choose 2 = 6 ways to choose the two kids who receive all 3 erasers. For each of these cases, we are left with 1 eraser to distribute among the remaining 2 kids. Using stars and bars, the number of ways to distribute the remaining eraser is 3 choose 1 = 3.

Case 3: Three kids receive all 3 erasers.
There are 4 choose 3 = 4 ways to choose the three kids who receive all 3 erasers. In this case, the remaining kid receives no eraser.

Case 4: All four kids receive all 3 erasers.
There is only one way for this case.

Step 4: Final count
To get the final count, we subtract

16