In how many different ways can we arrange 120 students into 6 groups for 6 different classes so that the largest group has at most 2 members more than the smallest group?

My initial plan was to use a generating function, but I stumbled across a problem. Let's mark the groups with numbers 1 to 6 and let ${n}_{i},i\in \{1,\dots ,6\}$ denote the number of members of the i-th group in some arrangment. To see where this would lead me, for a moment, I assumed ${n}_{1}\le {n}_{2}\le \cdots \le {n}_{6}\le {n}_{1}+2$ in hope to find some range $\{m,\dots M\}$ for ${n}_{i}$'s and use a generating function $f(x)=({x}^{m}+\cdots +{x}^{M}{)}^{6}$ and find $\u27e8{x}^{120}\u27e9-$ the coefficient in front of ${x}^{120}$, however students are distinct entities and m and M still remained misterious. I then tried figuring out if I was on the somewhat right track by, again taking $m=min\{{n}_{1},\dots ,{n}_{6}\}$ and write ${n}_{i}=m+{j}_{i},{j}_{i}\in \{0,1,2\}.$. I believe, an arrangement with 2 groups of 19,2 groups of 20 and 2 groups of 21 people suggests there should be at least 19 people in each group.