Eight-member team

We play a golf tournament where 4 pairs from team A will always play against 4 pairs from team B. Each team thus has in total 8 members. We tried to find out how many possible combinations there are of 4 playing groups, where in each there are 2 pairs – from each of the eight-member teams one?

How many are there then possible variants, if it also mattered which group the players go in – the foursomes start one after another in sequence.

It got us a little confused, so we would be glad for help.

Final Answer:

a =  28
b =  576

Step-by-step explanation:

$$\begin{gathered} C_2(8)=\left(\genfrac{}{}{0ex}{}{8}{2}\right)=\frac{8!}{2!(8-2)!}=\frac{8\cdot 7}{2\cdot 1}=28 \\ a=\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28 \end{gathered}$$

$$\begin{gathered} n=4\cdot 3\cdot 2\cdot 1=24 \\ b=n^2=24^2=576 \end{gathered}$$

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