To clarify my doubt over my wrong approach in combination question

Let's say the balls are numbered. Red balls from $1$ to $5$, and white balls from $1$ to $6$.

In your approach, you first pick two red balls and two white balls, then two balls freely. That means that you count

Red ball 1 and 2, then white ball 1 and 2, then red ball 3 and white ball 3

as a different choice from

Red ball 1 and 3, then white ball 2 and 3, then red ball 2 and white ball 1

and there are several other combinations that ought to be counted as the same choice as this one, but your method considers them different. In fact, there are 9 of them in total. If the final choice had 2 whites or 2 reds, that number would instead be 6.

This is why you get an answer that is much too large (somewhere between 6 and 9 times too large).

Your approach overcounts selections. Consider the following two scenarios:

• you take red balls 1 and 2 and white balls 1 and 2, then red balls 3 and 4
• you take red balls 1 and 3 and white balls 1 and 2, then red balls 2 and 4

These are in fact the same selection, but your approach counts them as different, so it overestimates the answer.