Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
First we’ll pick the team containing the youngest of the $33$ men; there are $\binom{32}{10}$ ways to do that, since we need only choose his $10$ teammates. Then we choose the team containing the youngest of the remaining $22$ men; this can be done in $\binom{21}{10}$ ways. That leaves $11$ men to form the third team. This approach does not overcount: there’s always one team that contains the youngest man and one that contains the youngest man not on that first team. Thus, there are
$$\binom{32}{10}\binom{21}{10}\tag{2}$$
ways to choose the three teams.
Of course it would be a good idea to make sure that $(1)$ and $(2)$ actually yield the same result:
$$\begin{align*}
\frac16\binom{33}{11}\binom{22}{11}&=\frac16\cdot\frac{33!}{11!22!}\cdot\frac{22!}{11!11!}\\\\
&=\frac13\cdot\frac{33\cdot32!}{11!22!}\cdot\frac12\cdot\frac{22\cdot 21!}{11!11!}\\\\
&=\frac{11\cdot32!}{11!22!}\cdot\frac{11\cdot21!}{11!11!}\\\\
&=\frac{32!}{10!22!}\cdot\frac{21!}{10!11!}\\\\
&=\binom{32}{10}\binom{21}{10}\;.
\end{align*}$$
You are correct in saying that the first person can be chosen in $6$ ways, the second in $5$ & the third in $4$ ... BUT
This group of $3$ people could have been chosen $3!$ ways ... SO
there will be $6 \times 5 \times 4 /3!$ ways.
Another way to think about it is ... how many ways could you choose $3$ people from $6$ ... $\binom{6}{3} = \color{red}{20}$.
Best Answer
You're on the right track.
The number of ways to pick Team 1 is equal to $\binom{40}6\cdot\binom{34}4$. After that, the number of ways to pick out Team 2 is $\binom{30}6\cdot \binom{24}4$. And so on.
Ultimately, the number of ways to pick out the full $4$ teams is $$ \binom{40}6\cdot\binom{34}4\cdot\binom{30}6\cdot\binom{24}4\cdot\binom{20}6\cdot\binom{14}4\cdot\binom{10}6\cdot\binom{4}4\\ = \frac{40!}{(6!)^4\cdot (4!)^4} $$ (Also known as the multinomial coefficient $\binom{40}{6, 4, 6, 4, 6, 4, 6, 4}$.)
However, we don't care which team is Team 1 and which team is Team 4. We just care which four teams are picked. Since the same four teams can be picked out in $4!$ ways, the total number of different team compositions is $$ \frac{40!}{(6!)^4\cdot (4!)^5} $$