The question asks to determine the amount of double matches given 8 players.

My approach: a double match consists out of 4 players. The number of ways to choose a team of 4 players equals $$\binom{8}{4} = 70.$$ Now we have two groups of $4$ players each. In each group, we select two players to form a team. In each group, this is done in $\binom{4}{2} = 6$ ways, hence the total amount of double matches equals $$70 \cdot 6 \cdot 6 = 2520.$$

However, the answer in my book says that the result is $210$ (and I feel that my answer is indeed way too much). What did I do wrong (and what possibilities do I double count with this reasoning?)

EDIT: I now see that I take the order of the two groups of 4 players into account. Hence I should divide my original answer by $2$. This gives $1260$, but this is still $6$ times the correct solution.

EDIT 2 New attempt on this question, given the answer of Aman: We pick 4 players from the 8 players. This gives us $$\binom{8}{4} = 70$$ possibilities. From these $4$ players, we need to pick $2$ to form the first team, the remaining $2$ form the second team. This can be done in $$\binom{4}{2} = 6$$ ways. However, the order in which the teams are formed does not matter, so there are only $6/2 = 3$ ways, giving a total of $70 \cdot 3 = 210$ ways.

  • $\begingroup$ Why did you multiply by 6 twice? Defining one doubles team automatically defines the other $\endgroup$ – aman Apr 8 at 16:59
  • $\begingroup$ @aman I choose 2 people in the first group of 4 and did the same in the second group. Even without this extra factor 6, I still double count some combinations, since the answer is half of the answer I would obtain if I divide mine by 6. $\endgroup$ – Student Apr 8 at 17:00

How many ways can I choose 2 people to form the first team from the original 8?

Answer: $8\choose2$ = 28

How many ways can I choose 2 people to form the second team from the remaining 6?

Answer: $6\choose2$ = 15

Divide by 2 as the order of selection does not matter

$\frac{28*15}{2} = 210$

  • $\begingroup$ What happens with the remaining 4 players? They could play multiple double matches, no? $\endgroup$ – Student Apr 8 at 17:09
  • $\begingroup$ You seem to misunderstand the question (or maybe I do) $\endgroup$ – aman Apr 8 at 17:09
  • $\begingroup$ How many ways can a single double's match be selected from a group of 8 players? $\endgroup$ – aman Apr 8 at 17:10
  • $\begingroup$ Not everyone has to play. $\endgroup$ – aman Apr 8 at 17:10
  • $\begingroup$ Sorry, I was confused. I see my error now (and also the error I made in my reasoning). I feel so stupid right now haha $\endgroup$ – Student Apr 8 at 17:10

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