Your answer of $6^4$ would be correct if we were considering ordered outcomes, where it matters what each die shows in order. In this scenario, you'd treat $1,1,1,2$ as different from $1,2,1,1$ for instance.
Your professor is considering unordered outcomes, which would treat both of the examples above as $3$ ones plus $1$ two.
One way to count the unordered outcomes is a stars and bars approach - Wikipedia. Here you imagine your 4 dice as 'stars' and your $6$ individual outcomes (numbers $1\dots6$) as $5$ 'bars'. An outcome looks something like:
$$\text{ones}\left|\vphantom\int\right.
\text{twos}\left|\vphantom\int\right.
\text{threes}\left|\vphantom\int\right.
\text{fours}\left|\vphantom\int\right.
\text{fives}\left|\vphantom\int\right.
\text{sixes}$$
The combination $1,2,1,1$ (3 ones, 1 two) would look like:
$$***\left|\vphantom\int\right.
*\left|\vphantom\int\right.
\left|\vphantom\int\right.
\left|\vphantom\int\right.
\left|\vphantom\int\right.
$$
The combination $2,4,4,5$ (1 two, 2 fours, 1 five) would look like:
$$\left|\vphantom\int\right.
*\left|\vphantom\int\right.
\left|\vphantom\int\right.
**\left|\vphantom\int\right.
*\left|\vphantom\int\right.
$$
All together there are 9 objects (4 stars, 5 bars), and any combination amounts to picking 5 of the locations for the bars - $\dbinom95$ possibilities, or picking 4 locations for the stars - $\dbinom94$ possibilities. Happily, they give the same result.
There are $12$, not $6$ ways to get a permutation of $6,6,5,4$. You can choose the position of the $5$ in $4$ ways, then choose the position of the $4$ in $3$ ways.
This gives us $4+12+4 = 20$ ways to get a total of $21$, and $4+6=10$ ways to get a total of $22$, and so we conclude that rolling a $21$ is twice as likely.
Best Answer
An outcome here is the same as a six-tuple of non-negative integers that sum to $6$, the $i^{th}$ entry telling you how many times $i$ came up as a value.
Stars and Bars tell us that the number of such is $$\binom {6+6-1}6=462$$