Solution:

Let $r$, $b$, $w$, and $y$ denote the number of red, blue, white, and yellow chips selected, respectively. We want the number of integer solutions to
$$r + b + w + y = 9$$
subject to $0 \leq r \leq 3$, $0 \leq b \leq 3$, $0 \leq w \leq 3$, $0 \leq y \leq 3$.

Since there are only 3 of each color, the maximum for each variable is 3.

We can use generating functions. The generating function for each color is $(1 + x + x^2 + x^3)$. The generating function for the total is
$$(1 + x + x^2 + x^3)^4$$
We want the coefficient of $x^9$ in this expansion.

Let us expand $(1 + x + x^2 + x^3)^4$.

Alternatively, note that $r + b + w + y = 9$ and $0 \leq r, b, w, y \leq 3$.

Let us count the number of non-negative integer solutions to $r + b + w + y = 9$ with $r, b, w, y \leq 3$.

We can use the inclusion-exclusion principle.

Let $N$ be the number of non-negative integer solutions to $r + b + w + y = 9$ with $r, b, w, y \leq 3$.

Let $S$ be the set of all non-negative integer solutions to $r + b + w + y = 9$.

The number of non-negative integer solutions is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} = 220$.

Now, subtract the solutions where at least one variable exceeds 3.

Let $A$ be the set where $r \geq 4$, $B$ where $b \geq 4$, $C$ where $w \geq 4$, $D$ where $y \geq 4$.

By inclusion-exclusion:

$N = S - (|A| + |B| + |C| + |D|) + (|A \cap B| + |A \cap C| + |A \cap D| + |B \cap C| + |B \cap D| + |C \cap D|) - (|A \cap B \cap C| + |A \cap B \cap D| + |A \cap C \cap D| + |B \cap C \cap D|) + |A \cap B \cap C \cap D|$

First, $|A|$ is the number of solutions with $r \geq 4$. Let $r' = r - 4 \geq 0$, so $r' + b + w + y = 5$. The number of non-negative integer solutions is $\binom{5 + 4 - 1}{4 - 1} = \binom{8}{3} = 56$.

Similarly, $|B| = |C| = |D| = 56$.

Now, $|A \cap B|$ is the number of solutions with $r \geq 4$ and $b \geq 4$. Let $r' = r - 4 \geq 0$, $b' = b - 4 \geq 0$, so $r' + b' + w + y = 1$. The number of non-negative integer solutions is $\binom{1 + 4 - 1}{4 - 1} = \binom{4}{3} = 4$.

There are $\binom{4}{2} = 6$ such pairs, so the sum over all pairs is $6 \times 4 = 24$.

Now, $|A \cap B \cap C|$ is the number of solutions with $r, b, w \geq 4$. Let $r' = r - 4$, $b' = b - 4$, $w' = w - 4$, so $r' + b' + w' + y = 9 - 4 \times 3 = -3$. There are no non-negative integer solutions to this equation, so $|A \cap B \cap C| = 0$.

Similarly, all triple and quadruple intersections are zero.

So,
$$N = 220 - 4 \times 56 + 6 \times 4 = 220 - 224 + 24 = 20$$

Therefore, there are $\boxed{20}$ ways to select nine chips from the bag.