Note [2020-12-25]: This is a completely revised answer.
This answer consists of three parts. In the first part we give a solution to the problem. The second part is a generalisation which can be derived easily and we find a nice identity of Stirling numbers of the second kind. In the third part we calculate a small example by hand as plausibility check.
A solution: This solution is based upon PIE, the principle of inclusion-exclusion.
The number of ways to distribute four identical balls to four persons so that each of them has zero or more balls is using stars and bars
\begin{align*}
\binom{4+4-1}{4-1}=\binom{7}{3}\tag{1}
\end{align*}
We start with a formula for distributing zero or more balls and not at least one ball, since balls with other colors will also be used for distribution to the people.
We have four different colors with four balls each. The number of ways to distribute these balls, so that each of the persons has zero or more balls is according to (1)
\begin{align*}
\binom{7}{3}^4\tag{2}
\end{align*}
Now we subtract from (2) according to PIE, the principle of inclusion-exclusion, the cases where at least one of the four persons has no ball. Since one person can be selected in $\binom{4}{1}$ ways, we get as number to subtract
\begin{align*}
\binom{4}{1}\binom{4+3-1}{3-1}=\binom{4}{1}\binom{6}{2}
\end{align*}
But here we did some overcounting, since we counted the cases where two of the four persons has no ball, twice. To compensate this we add $\binom{4}{2}$ times the number of ways to distribute the balls to two persons and so on. We obtain this way
\begin{align*}
\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^4\tag{3}
\end{align*}
Finally we have to respect that four red balls are not allowed to be distributed to one person only. In order to do so we calculate the number of ways to distribute three times four balls of the other colors to four persons. Since we have $\binom{4}{1}$ ways to give the four red balls to a person we have to subtract:
\begin{align*}
\binom{4}{1}\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^3\tag{4}
\end{align*}
One final aspect we missed so far is that we additionally have to subtract the $\binom{4}{1}=4$ cases where four red balls are given to a person and this person has no other balls. These $4$ cases also have to be subtracted.
We obtain from (3) and (4) the number of wanted ways as
\begin{align*}
&\color{blue}{\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^4-\binom{4}{1}\sum_{i=0}^3(-1)^i\binom{4}{i}\binom{7-i}{3-i}^3-\binom{4}{1}}\\
&\quad=\left[\binom{4}{0}\binom{7}{3}^4-\binom{4}{1}\binom{6}{2}^4+\binom{4}{2}\binom{5}{1}^4-\binom{4}{3}\binom{4}{0}^4\right]\\
&\quad\quad-\binom{4}{1}\left[\binom{4}{0}\binom{7}{3}^3-\binom{4}{1}\binom{6}{2}^3
+\binom{4}{2}\binom{5}{1}^3-\binom{4}{3}\binom{4}{0}^3\right]-\binom{4}{1}\\
&\quad=\left[1\cdot1\,500\,625-4\cdot50\,625+6\cdot625-4\cdot1\right]\\
&\quad\quad-4\left[1\cdot 42\,875-4\cdot 3\,375+6\cdot125-4\cdot1\right]-4\\
&\quad=1\,301\,871-120\,484-4\\
&\,\,\quad\color{blue}{=1\,181\,383}\tag{5}
\end{align*}
A generalisation:
We can easily derive the following generalisation of (3) from the approach above: We consider $n$ pairwise distinct colors $C_1, C_2, \ldots, C_n$ and $b_1+b_2+\cdots+b_n$ balls where $b_j$ balls have color $C_j, 1\leq j\leq n$. The number of ways to distribute $b_1+\cdots+b_n$ balls to $k$ persons is
\begin{align*}
\color{blue}{\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}\prod_{j=1}^n\binom{b_j+k-i-1}{k-i-1}}\tag{6}
\end{align*}
We consider the special case $b_1=b_2=\cdots=b_n=1$. Here we have $n$ different(ly colored) balls and we can reformulate the wanted number as $k!$ times the number of ways to partition $n$ numbers into $k$ sets. This is ${n \brace k}k!$, with $n\brace k$ the well-known Stirling numbers of the second kind.
We derive from (6) the identity
\begin{align*}
\color{blue}{{n\brace k}k!}&=\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}\prod_{j=1}^n\binom{1+k-i-1}{k-i-1}\\
&=\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}\binom{k-i}{k-i-1}^n\\
&\,\,\color{blue}{=\sum_{i=0}^{k-1}(-1)^i\binom{k}{i}(k-i)^n}\\
\end{align*}
An example:
Here we consider as application and plausibility check of the formulas (5) and (6) the following problem.
- Find the number of ways to distribute $2$ identical red balls , $2$ identical green balls, $2$ identical white balls among two people so that each person receives at least one ball but no one gets $2$ red balls.
We have the situation $n=3, b_1=b_2=b_3=2$ and we have two persons, which means $k=2$. The wanted number is following (5) and (6)
\begin{align*}
&\sum_{i=0}^{2-1}(-1)^i\binom{2}{i}\binom{2+2-i-1}{2-i-1}^3\\
&\qquad-\binom{2}{1}\sum_{i=0}^{2-1}(-1)^i\binom{2}{i}\binom{2+2-i-1}{2-i-1}^2-2\\
&=\sum_{i=0}^{1}(-1)^i\binom{2}{i}\binom{3-i}{1-i}^3
-2\sum_{i=0}^{1}(-1)^i\binom{2}{i}\binom{3-i}{1-i}^2-2\\
&=\left[\binom{2}{0}\binom{3}{1}^3-\binom{2}{1}\binom{2}{0}^3\right]
-2\left[\binom{2}{0}\binom{3}{1}^2-\binom{2}{1}\binom{2}{0}^2\right]-2\\
&=25-2\cdot7-2\\
&\,\,\color{blue}{=9}
\end{align*}
We list all $25$ cases, give the $9$ valid cases in boldface and mark the $2\cdot7$ and $2$ cases which are not valid blue and red. We have
\begin{align*}
\begin{array}{cccccc}
\mathbb{g|g\color{blue}{rr}ww}&\mathbb{gg|\color{blue}{rr}ww}&\mathbf{ggr|rww}&\mathbb{gg\color{blue}{rr}|ww}&\mathbb{gg\color{blue}{rr}w|w}&\mathbf{ggrw|rw}\\
\mathbf{ggrww|r}&\mathbb{ggw|\color{blue}{rr}w}&\mathbb{ggww|\color{red}{rr}}&\mathbf{ggr|grww}&\mathbb{g\color{blue}{rr}|gww}&\mathbb{g\color{blue}{rr}w|gw}\\
\mathbb{g\color{blue}{rr}ww|g}&\mathbf{grw|grw}&\mathbf{grww|gr}&\mathbb{gw|g\color{blue}{rr}w}&\mathbb{gww|g\color{blue}{rr}}&\mathbf{r|ggrww}\\
\mathbb{\color{red}{rr}|ggww}&\mathbb{\color{blue}{rr}w|ggw}&\mathbb{\color{blue}{rr}ww|gg}&\mathbf{rw|ggrw}&\mathbf{rww|ggr}&\mathbb{w|gg\color{blue}{rr}w}\\
\mathbb{ww|gg\color{blue}{rr}}&&&&&
\end{array}
\end{align*}
