Because B played 2 1 and A only played 10, because as long as you lose, you have to come down, and the next set must be played by the idle side, so any party of ABC plays at least one of any set or two. Because A only played 10, and B played twice as many times as A, it can be seen that B participated in all 2 1 games and won the first 20 games. Only in this way can B play 2 1 game.

And BC played first, so A played the 2nd, 4th, 6th, 8th, 10, so C played the 3rd, 5th, 7th and 9th.

That is to say, even the disk under AB; BC under the odd disk;

So BC downloaded 9 sets.

Supplement (interesting inference): As for the result of 2 1 set BC, it's hard to say. Of course, B is so powerful that it is likely that B will win. So the number of winners and losers may be one: A 10 war 10 loss; B 2 1 Zhan 2 1 win; C 1 1 war 1 1 defeat

There may be two winners and losers: A 10 and 10 losses; B 2 1 20 wins 1 negative; C 1 1 war 1 victory 10 negative