Define $f(n, k)$ as the number of ways to distribute $k$ chocolates to $n$ children such that each child receives 0, 1, or 2 chocolates. For example, $f(3,4)=6$, $f(3,6)=1$, and $f(3,7)=0$.

(a) List all 6 ways to distribute 4 chocolates to 3 children with each one receiving at most two chocolates.

(b) Considering 2015 children, verify that $f(2015, k)=0$ for all $k$ greater than or equal to an appropriate value.

(c) Show that the equation

$$
f(2016, k)=f(2015, k)+f(2015, k-1)+f(2015, k-2)
$$

is true for all positive integers $k$ greater than or equal to 2.

(d) Calculate the value of the expression

$$
f(2016,1)+f(2016,4)+f(2016,7)+\ldots+f(2016,4027)+f(2016,4030)
$$
Solution

(a) Let's represent each distribution by an ordered triple of numbers $(a, b, c)$, where each number represents the amount of chocolates each child will receive. The six possibilities are:

$$
(2,2,0) ;(2,0,2) ;(0,2,2) ;(2,1,1) ;(1,2,1) ;(1,1,2)
$$

(b) If $k \geq 2 \cdot 2015+1=4031$, then by the Pigeonhole Principle, if $k$ chocolates are distributed to 2015 children, at least one of them will receive more than 2 chocolates. In other words, it is impossible for each to receive at most two chocolates. Therefore, for $k \geq 4031$, we have $f(2015, k)=0$.

(c) Let's consider the possibilities of chocolates for the first child. If she gets 0, then there are $k$ chocolates left for the other 2015. If she gets 1, there are $k-1$ left for the others. And if she gets 2, then there are $k-2$ left for the rest. This count in three cases corresponds to the following equation:

$$
f(2016, k)=f(2015, k)+f(2015, k-1)+f(2015, k-2)
$$

(d) We will call this sum $S$. Using the previous item, we have

$$
\begin{aligned}
f(2016,1) & =f(2015,1)+f(2015,0) \\
f(2016,4) & =f(2015,4)+f(2015,3)+f(2015,2) \\
f(2016,7) & =f(2015,7)+f(2015,6)+f(2015,5) \\
& \cdots \\
f(2016,4027) & =f(2015,4027)+f(2015,4026)+f(2015,4025) \\
f(2016,4030) & =f(2015,4030)+f(2015,4029)+f(2015,4028)
\end{aligned}
$$

Adding everything, we will have

$$
S=f(2015,0)+f(2015,1)+\ldots+f(2015,4029)+f(2015,4030)
$$

Notice that the second term represents all the ways to distribute 0, 1, or 2 chocolates to each child in a group of 2015 children. Therefore, using the multiplication principle, this sum is equal to $3^{2015}$.