There are 10 people in the club. I want to make 3 teams out of them. Two of the teams (team A and B) will have 3 good players, and the latter one (team C) will have the 4 remaining students. In how many ways can I choose these teams?


The answer depends very much on how many good players there are. The simplest assumption is that there are exactly $6$.

Imagine first that we will give the teams labels A, B, C, maybe t-shirts with these letters on them. Then there are $\dbinom{6}{3}$, that is, $20$ ways to select the people who will wear the A shirts, and now the rest of the division into teams is determined.

However, if we are just thinking about division into unlabelled teams, when shirts are removed the division that has say players $1$, $2$, $3$ in Team A, and $4$, $5$, $6$ in Team B is indistinguishable from the division where we have $4$, $5$, $6$ in Team A and $1$, $2$, $3$ in Team B. So there are only $\frac{20}{2}$ ways of dividing our group into unlabelled teams.

Similar reasoning takes care of more complicated cases where there are more than $6$ good players. For example, suppose there are $7$. Then there are $\dbinom{7}{1}$ ways to decide which good player will go to the team of four. For every such way, there are, for labelled teams, $\dbinom{6}{3}$ ways to decide who goes to Team A, and we get a total of $\dbinom{7}{1}\dbinom{6}{3}$. For unlabelled teams, divide by $2$.

  • $\begingroup$ So, if all players are candidates to be good players, the solution should be $\dbinom{10}{4}\dbinom{6}{3} = 4200$? $\endgroup$ – Helen Nov 25 '12 at 18:36
  • $\begingroup$ @Helen: If all players are good players, and we are dividing into labelled teams, your expression is right. As before, for unlabelled, divide by $2$. $\endgroup$ – André Nicolas Nov 25 '12 at 18:39
  • $\begingroup$ Thank you very much for the detailed answer. $\endgroup$ – Helen Nov 25 '12 at 18:45

i think:

$(10 \cdot 9 \cdot 8) \cdot (7 \cdot 6 \cdot 5)=151200$

  • $\begingroup$ I would be happy if you add some explanation. It looks like your answer is $10!/7! * 7!/4!$... but why? $\endgroup$ – Helen Nov 25 '12 at 18:23
  • $\begingroup$ for example: for first Person team A has 10 choise, for second person team A or B have 9 choise, for six person team A or B have 5 choise. and last 4 person for team C. $\endgroup$ – Mj125 Nov 25 '12 at 18:27
  • $\begingroup$ I think the order is important in your answer, but it shouldn't be. There are $3!$ choices that make team A from the same 3 people. $\endgroup$ – Helen Nov 25 '12 at 18:38

Great question;

looks like you are looking for how many combinations have can have dividing $10$ people as $4$, $3$, and $3$ again;

Very easy to see visually using factorials


Expanding out completely:

$$ \frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{4\times3\times2\times1\times3\times2\times1\times3\times2\times1}$$

Canceling first group out: $$\frac{10\times9\times8\times7\times6\times5}{3\times2\times3\times2}$$

Canceling second group out: $$\frac{10\times9\times8\times7\times5}{3\times2}$$

Canceling third group out: $$5\times3\times8\times7\times5 = 4200$$


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