Casework and another argument yielding contradictory results I am trying to solve this counting problem which asks to find the number of ways that you can select 6 balls out of 5 distinct red balls and 6 distinct white balls, with atleast 2 balls of each color.
My book claims that it is necessary to do this by casework and gives an answer 425 with the following consideration:
$$\textrm{Number of ways}= \binom{5}{2}\binom{6}{4}+\binom{5}{3}\binom{6}{3}+\binom{5}{4}\binom{6}{2}=425 $$
And I understand that well enough.

I came up with the following way. 
There are $\binom{5}{2}$ ways to choose the minimum 2 balls from red and $\binom{6}{2}$ ways to choose the minimum 2 white balls.
Finally from the remaining 7 balls we pick two balls to finish of.
$$\textrm{Number of Ways}=\binom{5}{2}\binom{6}{2}\binom{7}{2}=21\cdot 15\cdot 10=3150$$

I have a feeling that I'm grossly overcounting and that there are arrangements where I say, select 2 red balls before and 2 after which is the same as another in which I choose the latter 2 before.
Is there a way to salvage this and/or another approach that is more elegant than casework?
 A: 3red/3white are counted $3\times 3=9$ times as in your question & the comment. In detail: number the balls $R_1$ to $R_5$ and $W_1$ to $W_6$. Each of the following 9 cases are counted differently in your calculation when they should really be counted once:
$$\{R_1, R_2\}, R_3; \{W_1, W_2\}, W_3\\
\{R_1, R_2\}, R_3; \{W_1, W_3\}, W_2\\
\{R_1, R_2\}, R_3; \{W_2, W_3\}, W_1\\
\{R_1, R_3\}, R_2; \{W_1, W_2\}, W_3\\
\{R_1, R_3\}, R_2; \{W_1, W_3\}, W_2\\
\{R_1, R_3\}, R_2; \{W_2, W_3\}, W_1\\
\{R_2, R_3\}, R_1; \{W_1, W_2\}, W_3\\
\{R_2, R_3\}, R_1; \{W_1, W_3\}, W_2\\
\{R_2, R_3\}, R_1; \{W_2, W_3\}, W_1$$
There's nothing special about $R_1,R_2,R_3,W_1,W_2,W_3$ here; same holds for all pairs of 3red/3white.
4red/2white and 2red/4white are each counted $\binom{4}{2}=6$ times in your count (there's $\binom{4}{2}$ different ways to split $\{R_1, R_2, R_3, R_4\}$ into 2 sets of 2: the set of 2 picked before & the set of 2 picked after).
I can't really see any way other than to separate these two cases, which leads to a solution probably less elegant than the casework, but it's maybe good to see how you get to the same answer.
There's $\binom{5}{3} \binom{6}{3} = 200$ ways to pick 3/3.
So using the correcting-for-overcounting approach we get that the total # ways to pick 3r/3w OR 4r/2w, 2r/4w is
$$200 + \frac{3150-200\times 9}{6} = 200 + 225 = 425.$$
Can't think of some other more elegant solution.
