28 must be wrong. Imagine that there are only four teams in ABCD, and there are only three ways to divide them into two groups.
( 1)AB CD (2)AC BD (3) AD BC
Instead of c (4,2) = 6 kinds. ..
therefore
C (8,2) Calculate how many pairs of different combinations a * * * can draw. ..
But it's not just a matter of choosing one group, it's divided into four groups, and the four groups and eight teams are different.
There are only 28 different groups, and there are 4 groups after each grouping.
So there are only seven ways. ..