I'll reword your question a bit. Let me know if I have not framed it properly.
Say we have a collection of $r$ unique elements each of which occurs in multiple copies. Namely, our collection has $n_1$ identical copies of $X_1$, $n_2$
identical copies of $X_2$, $\ldots$, and $n_r$ identical copies of
$X_r$.
How many ways can we form combinations of length $k$, where repeated copies of unique elements are allowed?
This question has a standard answer which can be derived informally and also formally (see below). In either case the answer is
$$
\sum_{\ell=0}^{r} (-1)^{\ell} \sum_{\textbf{j} \in \{n_k\}_{\ell} } \binom{k+r-1 - \sum_{i=1}^{\ell}(j_i +1)}{r-1}
$$
where $\{n_k\}_{\ell}$ is the set of all $\ell$-length combinations of the elements $(n_1, \ldots, n_r)$, $\textbf{j} = (j_1, j_2, \ldots, j_{\ell})$ is a vector representing a particular element of the set of $\{n_k\}_{\ell}$, and the second summation is over all such combinations. (The $\ell=0$ term simply reduces to "$k+r-1$ choose $r-1$".)
We can apply the above formula to your example. In your case you have $r=3$ types of elements and you're trying to form combinations of length $k=4$. We identify the blue, orange, and yellow balls with $X_1$, $X_2$, and $X_3$ respectively. We then have $n_1 = 3$, $n_2 = 1$, and $n_3 =4$. The possible values of $\textbf{j}$ with defined by various $\ell$ are
\begin{eqnarray}
\ell = 1: & \quad j_1 & \in \{3, 1, 4\} \\
\ell = 2: & \quad (j_1, j_2) & \in \{(3, 1), (1,4), (4, 3)\} \\
\ell = 3: & \quad (j_1, j_2, j_3) & \in \{(3, 1, 4) \}
\end{eqnarray}
We thus find
\begin{eqnarray}
& & \binom{4+3-1}{3-1} - \Big[ \binom{4+3-1 -(3+1)}{3-1} +\binom{4+3-1 -(1+1)}{3-1}+\binom{4+3-1 -(4+1)}{3-1}\Big] \\
& & + \Big[ \binom{4+3-1 -(4+2)}{3-1} +\binom{4+3-1 -(5+2)}{3-1}+\binom{4+3-1 -(7+2)}{3-1}\Big]\\
& &- \Big[ \binom{4+3-1 -(8+3)}{3-1}\Big]\\
& = & \binom{6}{2} - \Big[\binom{2}{2} + \binom{4}{2}+ 0\Big] +\Big[0\Big] - 0 = 8,
\end{eqnarray}
where we used $\binom{n}{k} = 0$ if $n>k$. This result can be checked by listing the number of ways to form the elements.
Formal Derivation
To derive the general answer to the highlighted question, we note that the desired number is equal to the number of ways $\alpha_1$, $\alpha_2, \ldots, \alpha_r$ can all sum to $k$ where each $\alpha_k$ (i.e., the number of elements of $X_k$) runs from $0, 1, \ldots, n_k$. Namely, we want to compute
$$
(\#)_{n_1,\ldots, n_r; k} = \sum_{\alpha_1=0}^{n_1}\cdots \sum_{\alpha_r=0}^{n_r} \delta_{k, \alpha_1+\cdots +\alpha_r},
$$
where $\delta_{\ell, m}$ is the Kronecker delta. To compute the above quantity, we need to use two identities both of which are established by the Cauchy's Integral Formula:
$$
\delta_{m, \ell} = \frac{1}{2\pi i} \oint_C \frac{dz}{z} z^{m-\ell}, \qquad \binom{k}{\ell} = \frac{1}{2\pi i} \oint_C dz\, \frac{z^{k}}{(z-1)^{\ell +1}}.
$$
Using the contour integral representation of the Kronecker delta, we have
\begin{align}
(\#)_{n_1,\ldots, n_r; k} & = \sum_{\alpha_1=0}^{n_1} \cdots \sum_{\alpha_{r}=0}^{n_r} \oint_C \frac{dz}{z} z^{k- \alpha_1- \cdots - \alpha_r}\\
& = \oint_C \frac{dz}{z} z^{k} \prod_{m=1}^{r} \sum_{\alpha_m=0}^{n_m} z^{-\alpha_k} \quad \text{[Commute summation and integral]} \\
& = \oint_C \frac{dz}{z} z^{k} \prod_{m=1}^{r} \frac{1-z^{-n_m-1}}{1-z^{-1}} \quad \text{[Geometric Series Identity]} \\
& = \oint_C dz \frac{z^{k+r-1}}{(z-1)^r} \prod_{m=1}^{r} \left(1- z^{-1} z^{-n_k}\right).
\end{align}
Now, using the identity
\begin{equation}
\prod_{i=1}^N(1+\lambda x_i) = \sum_{\ell=0}^{N} \lambda^{\ell} \Pi_{\ell}(x_1, \ldots, x_N),
\end{equation}
where $\Pi_{\ell}(x_1, \ldots, x_N)$ is the $\ell$th elementary symmetric polynomial in the variables $(x_1, \ldots, x_N)$, we find
\begin{align}
(\#)_{n_1,\ldots, n_r; k}
& = \sum_{\ell=0}^{r}(-1)^{\ell}\,\oint_C dz \frac{z^{k+r-1- \ell}}{(z-1)^r} \, \Pi_{\ell}(z^{-n_1}, \ldots, z^{-n_r}).
\label{eq:omegpre}
\end{align}
For an arbitrary elementary symmetric polynomial we have the definition
\begin{equation}
\Pi_{\ell}(x_1, x_2, \ldots, x_r) = \sum_{ \textbf{j} \in \{1, 2, \ldots, r\}_{\ell} } x_{j_1} x_{j_2}\cdots x_{j_{\ell}},
\end{equation}
where $\textbf{j} = (j_1, \ldots, j_{\ell})$ is a particular combination of length $\ell$ of the elements in $\{1, 2, \ldots, r\}$ and the summation is over all combinations of length $\ell$. For our case, we similarly have
\begin{equation}
\Pi_{\ell}(z^{-n_1}, \ldots, z^{-n_r}) = \sum_{ \textbf{j} \in \{n_k\}_{\ell} } z^{- j_1 - \ldots - j_{\ell}}.
\end{equation}
$(\#)_{n_1,\ldots, n_r; k} $ therefore becomes
\begin{align}
(\#)_{n_1,\ldots, n_r; k}
& = \sum_{\ell=0}^{r}(-1)^{\ell}\, \sum_{ \textbf{j} \in \{n_m\}_{\ell} }\oint_C dz \, \frac{z^{k+r-1- \ell}}{(z-1)^r} \,z^{- j_1 - \ldots - j_{\ell}} \\
& = \sum_{\ell=0}^{r}(-1)^{\ell}\, \sum_{ \textbf{j} \in \{n_m\}_{\ell} }\oint_C dz\, \frac{z^{k+r-1- \sum_{i=1}^{\ell}(j_i +1)}}{(z-1)^r}.
\end{align}
Using the contour integral representation of the binomial, we ultimately find
\begin{equation}
(\#)_{n_1,\ldots, n_r; k} = \sum_{\ell=0}^{r}(-1)^{\ell}\, \sum_{ \textbf{j} \in \{n_m\}_{\ell} } \binom{k+r-1 - \sum_{i=1}^{\ell}(j_i +1)}{r-1},
\label{eq:omeggen}
\end{equation}
where $j_1, \ldots, j_{\ell}$ is a particular combination of $\ell$ elements in $\{n_1, \ldots, n_r\}$ and the second summation runs over all such combinations.
Some Special Cases
For $n_k=1$ for all $k$, $\{n_1, \ldots, n_r\}$ becomes the $r$ element set $\{1, \ldots, 1\}$. Thus summing over all $\ell$-length combinations of $\{n_1, \ldots, n_r\}$ reduces to setting $j_i= 1$ for all $i$ and multiplying the relevant term by the number of ways to choose $\ell$ elements from a set of $r$ elements, i.e., $\binom{r}{\ell}$. We thus find
\begin{equation}
(\#)_{n_1,\ldots, n_r; k} = \sum_{\ell=0}^{r}(-1)^{\ell}\, \binom{r}{\ell} \binom{k+r-1 - 2\ell}{r-1}.
\end{equation}
However, if all $n_m$ are equal to 1, then the number of ways we can choose combinations of length $k$ from the elements $A_1, A_2, \ldots, A_r$ is simply the number of ways to choose $k$ elements from a set of $r$ elements. Thus we can conclude
\begin{equation}
\sum_{\ell=0}^{r}(-1)^{\ell}\, \binom{r}{\ell} \binom{k+r-1 - 2\ell}{r-1} = \binom{r}{k}.
\end{equation}
For $n_m = \infty$ for al $m$, then $j_i = \infty$ for all $i$ and all terms in \refew{omeggen} are zero except for $\ell=0$. (The binomial $\binom{n}{k}$ is zero whenever $n< k$). We thus have
\begin{equation}
(\#)_{n_1,\ldots, n_r; k} = \binom{k+r-1}{r-1},
\end{equation}
which matches the standard result (see "Combination with repetitions." for example).
For $n_m = n$ for all $m$, then $j_i = n$ for all $i$. Similar to the case in 1., summing over over all $\ell$-length combinations of $\{n_1, \ldots, n_r\}$ reduces to setting $j_i= n$ for all $i$ and multiplying the relevant term by the number of ways to choose $\ell$ elements from a set of $r$ elements. We therefore have
\begin{equation}
(\#)_{n_1,\ldots, n_r; k} = \sum_{\ell=0}^{r}(-1)^{\ell}\, \binom{r}{\ell} \binom{m+r-1 - \ell(n+1)}{r-1},
\end{equation}
which matches the expression in the intuitive argument given by Thoma in "Basic inclusion exclusion problem - picking 24 balls out of four sets of 10 balls".
