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.
A common way to explain it is known as "stars and bars".
I shall illustrate it putting identical balls into distinct bins ($1-6$) depicting the results obtained
One result with $n= 10$, say, would be $\;\;\bullet|\bullet|\bullet|\bullet\bullet\bullet\bullet|\bullet
|\bullet\bullet\;\;$
Make two notes: only $5$ dividers are needed to depict $6$ bins, and you could have $0$ balls in some bins, e.g. $||\bullet\bullet\bullet\bullet|\bullet\bullet\bullet|\bullet\bullet\bullet|$ depicting $0-0-4-3-3-0$
So if there are $n$ balls and $k$ bins ($k-1$ dividers), the only choice you have is to place the dividers among the lot, thus
$\dbinom{n+k-1}{k-1}$ which works out to $\dbinom{n+5}{5}$ for your particular example.
You could profitably look here to have a more detailed explanation
Best Answer
Use inclusion/exclusion principle:
So the total number of desired outcomes is:
$$\binom66\cdot6^8-\binom65\cdot5^8+\binom64\cdot4^8-\binom63\cdot3^8+\binom62\cdot2^8-\binom61\cdot1^8=191520$$