Why this solution does not works for the problem "Find number of ways to put 5 identical balls in two different bags"?

Question

I understand the bar and star method to solve this problem. There are 6 ways to put the balls into the bag.

But I was trying to figure out another approach to solve this question.

($5C0$ x $5C5$) + ($5C1$ x $4C4$) + ($5C2$ x $3C3$) + ($5C3$ x $2C2$) + ($5C4$ x $1C1$) + ($5C5$ x $0C0$)

i.e 
(choose 0 balls from 5 for bag1 AND choose all 5 balls from 5 for bag2) OR
(choose 1 balls from 5 for bag1 AND choose all 4 balls from 4 for bag2) OR
(choose 2 balls from 5 for bag1 AND choose all 3 balls from 3 for bag2) OR
(choose 3 balls from 5 for bag1 AND choose all 2 balls from 2 for bag2) OR
(choose 4 balls from 5 for bag1 AND choose all 1 balls from 1 for bag2) OR
(choose 5 balls from 5 for bag1 AND choose all 0 balls from 0 for bag2) OR

The experiment fails, of course.

You can laugh :) . But I am semi-beginner in this subject and wants to develop an intuition to solve this kind of problems. I understand how the star and the bar solution works. My solution which I provided must gives you a hint on how I think and have understood the concept. Can you teach me something which would help me to solve these kind of problems.

Answer 1

See, the balls are identical, so doing $5\choose0$,$5\choose1$,etc is meaningless because WHICH ball you choose doesn't matter In fact, you were on the right track.

There are six possibilities:

5 balls in first box and 0 in second

4 balls in first box and 1 in second

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0 balls in first box and 5 in second