<s>[INST] Counting the ways to form 4 different teams We have 40 players and we need to form 4 teams. For each team is necessary to indicate explicitly 6 players and 4 reserves. How many ways can teams be formed?
My attempt: first we can start to calculate in how many ways we can choose 6 players and 4 reserves. Since a distinction is made between players and reserves, I think we should consider it in the calculation (in the sense that, otherwise, it would have been simply 11 players). I think that we can calculate this number by multiplicate $\binom{40}{6}$ ways to choose the players and $\binom{34}{4}$ ways to choose the reserves. Now, the multiplication produces a very large number and makes me believe that I'm wrong (because then I need to calculate in how many ways we can assign one of the combinations to a team). Am I wrong? Is this a correct reasoning?
 [/INST] 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
$$
\frac{40!}{(6!)^4\cdot (4!)^5}
$$
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