To clarify my doubt over my wrong approach in combination question The question is:

A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least two balls of each colour.

I know this an easy question and we can easily do this by taking different cases and adding them like:-  $ { 5 \choose 2 }×{6\choose4}+  { 5 \choose 3}× { 6 \choose 3 }+ { 5 \choose 4 }× { 6 \choose 2 } =425$
But i want to know that why my first approach is wrong that is since we want two red balls and two white balls so we take  $ { 5 \choose 2 }×{6\choose2}$ and now the remaining two balls could be taken from the remaining 7 balls (11-4) so the final answer according to me -
$ { 5 \choose 2 }×{6\choose2}×{7\choose2}$ 
Please tell what is wrong with my approach I am very confused (the correct answer is 425).
 A: 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).
A: 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.
A: Let red balls are named as R1,R2,R3...R6 and white are named as W1 , W2, W3......W6.
In selection of 2 Red and 2 white let us say you have first got R1,R2,W1,W2 and later in next two you got R3 , W3 i.e you have now R1,R2,R3,W1,W2,W3 in 1800 ways.
This also includes when in first 4 you withdraw R1,R3,W1,W2 and next two are R2 , W3 now again you can see it is same set of balls but counted twice like wise there are many repetition of cases.
