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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?

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

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

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

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  • $\begingroup$ Wow, your method is much quicker than mine! +1 $\endgroup$ – Zubin Mukerjee Jan 18 at 12:43
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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

$$ \displaystyle\frac{ \displaystyle\binom{40}{16} \displaystyle\binom{24}{6} \displaystyle\binom{18}{6} \displaystyle\binom{12}{6} \displaystyle\binom{16}{4} \displaystyle\binom{12}{4} \displaystyle\binom{8}{4}}{24} $$

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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: $$4!\times\frac{40!}{4!^66!^4}=\frac{40!}{4!^56!^4}$$possibilities.

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