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?

Correct Option: A

Let the number of erasers received by the four kids be $x_1,x_2,x_3,$ and $x_4$.

We need to find the number of integer solutions to:
$x_1+x_2+x_3+x_4=7$
subject to the constraints: $1\le x_i\le 3$ for each $i\in \{1,2,3,4\}$.

Let us first distribute 1 eraser to each of the 4 kids to satisfy the "at least one" condition.
This uses up $4\times 1=4$ erasers, leaving $7-4=3$ erasers to distribute.

Let $y_i=x_i-1$ be the number of additional erasers kid $i$ gets.
The new equation is:
$y_1+y_2+y_3+y_4=3$
where $0\le y_i\le 2$ (since $x_i\le 3\Rightarrow y_i+1\le 3\Rightarrow y_i\le 2$).

We find the total number of non-negative integer solutions to $y_1+y_2+y_3+y_4=3$ using the stars and bars formula:
$\left(\genfrac{}{}{0ex}{}{3+4-1}{4-1}\right)=\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{6\times 5\times 4}{3\times 2\times 1}=20$ ways.

Now we subtract the cases that violate the upper bound condition $y_i\le 2$.
A violation occurs if any kid gets $y_i\ge 3$. Since the sum of all $y$ values is 3, the only way a kid can get 3 or more is if exactly one kid gets 3 and the rest get 0.
The possible violating distributions are:
(3, 0, 0, 0), (0, 3, 0, 0), (0, 0, 3, 0), and (0, 0, 0, 3).
There are exactly 4 such violating distributions.

Subtracting these violating cases from the total:
Number of valid ways = $20-4=16$.