Argthtjtr

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?

You're on the right track.

The number of ways to pick Team 1 is equal to $binom{40}6cdotbinom{34}4$. After that, the number of ways to pick out Team 2 is $binom{30}6cdot binom{24}4$. And so on.

Ultimately, the number of ways to pick out the full $4$ teams is
$$
binom{40}6cdotbinom{34}4cdotbinom{30}6cdotbinom{24}4cdotbinom{20}6cdotbinom{14}4cdotbinom{10}6cdotbinom{4}4\
= frac{40!}{(6!)^4cdot (4!)^4}
$$
(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
$$
frac{40!}{(6!)^4cdot (4!)^5}
$$

I would do it a little different. First split 40 players in to groups of 10 players, that we can do on $$a= {1over 4!} cdot {40choose 10}cdot {30choose 10}cdot {20choose 10}cdot {10choose 10}$$
ways.

Now in each group choose 4 reserve players, so we can do that on $b={10choose 4}^4$ ways.

Thus the result is $$ acdot b = {40!over 4!^5cdot 6!^4}$$

The number of ways to choose which players are reserves is $$binom{40}{16} $$

The remaining $24$ players are starters.

The number of ways to choose the $6$ starters on the first team is $$binom{24}{6} $$

Given this, the number of ways to choose the $6$ starters on the second team from the remaining $18$ choices is $$binom{18}{6}$$

Similarly, the third team's starters: $$binom{12}{6} $$

The remaining $6$ must start for the last team.

The total number of ways to arrange the starters within teams is

$$ binom{24}{6} binom{18}{6} binom{12}{6}$$

The same reasoning applies to reserves, and the total number of ways to arrange the reserves within teams is

$$ binom{16}{4} binom{12}{4} binom{8}{4}$$

Finally, the teams are not given distinguishing features (such as names) beforehand, so the order in which a set of four rosters is obtained is irrelevant. We've counted every assignment of the players $4!=24$ times instead of just once.

To compensate for this we can simply divide by $24$.

The final result is

$$ displaystylefrac{ displaystylebinom{40}{16} displaystylebinom{24}{6} displaystylebinom{18}{6} displaystylebinom{12}{6} displaystylebinom{16}{4} displaystylebinom{12}{4} displaystylebinom{8}{4}}{24} $$

There are: $$frac{40!}{4!^46!^4}$$ways to split up the $40$ players in $4$ groups of $4$ and $4$ groups of $6$ in such a way that every player is member of exactly one group, and ordering of the groups of $4$ as well as ordering of the groups of $6$ counts.

Then there are: $$frac1{4!}frac1{4!}frac{40!}{4!^46!^4}=frac{40!}{4!^66!^4}$$ways to split up the $40$ players in $4$ groups of $4$ and $4$ groups of $6$ in such a way that every player is member of exactly one group, and ordering of the groups of $4$ as well as ordering of the groups of $6$ does not count.

This determines $4$ selections and $4$ reserve-teams.

There are $4!$ ways to connect each selection with one reserve team.

That gives a total of: $$4!timesfrac{40!}{4!^66!^4}=frac{40!}{4!^56!^4}$$possibilities.