Selection of atleast 2 balls of each colour 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 atleast two balls
  of each colour.

I am trying to solve it as follows:
To guarantee $2$ red balls, choosing $2$ red balls : $\binom{5}{2}$.
To guarantee $2$ white balls, choosing $2$ white balls : $\binom{6}{2}$.
For a total of $6$ balls, choosing $2$ from the remaining $7$ balls : $\binom{7}{2}$
So total number of ways = $\binom{5}{2} \times\binom{6}{2} \times \binom{7}{2}$
However, the answer is $425$. What am I doing wrong and why?

Let's say red balls are $R_1$, $R_2$, $R_3$, $R_4$, $R_5$ and white balls are $W_1$, $W_2$, $W_3$, $W_4$, $W_5$, $W_6$. 
Now, assume we are choosing the balls as in your method and we have chosen $R_1$ and $R_2$ from red balls, $W_1$ and $W_2$ from white balls first and other two balls are $R_3$ and $W_3$. But this is as same as choosing $R_1$ and $R_3$ from red balls, $W_1$ and $W_3$ from white balls first and other two balls are $R_2$ and $W_2$. So we are overcounting.
In order to solve this, we should choose all required balls at once like the following:
If we have at least $2$ red balls and $2$ white balls, we can choose them as $2$ red $4$ white balls or $3$ red $3$ white balls or $4$ red $2$ white balls, which is $$\binom{5}{2} \binom{6}{4} + \binom{5}{3}\binom{6}{3}+\binom{5}{4} \binom{6}{2} = 150 + 200 + 75 = 425$$
