We have 8 balls, separating them into groups of size 3, we get 2 groups of size 3 and inevitably a group of size 2:
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I wanna see in how many ways is it possible to do so.
I choose the dividers (the lines) : $ 8-1 \choose 3-1$ $ = $ $ 7 \choose 2 $
But I was thinking, can I calculate the combination of choosing groups of 3 balls out of 8 balls to give me the answer? as followed: $ 8 \choose 3 $$ 8-3 \choose 3 $$ 8-3-3 \choose 2 $ = $ \frac {8!}{3!3!2!} $
But they are obviously different. I'm trying to see if it's the same if I group the balls by choosing them or group them choosing the dividers. they don't give same answer. why?