As the question follows,
In how many ways if we do not want $4$ cards of the same colour?
My solution so far
Total number of options of selecting 4 any cards: $495$
The number of options of $4$ of the same colour is: $175$
Therefore that means $495 - 175 = 320$ options that dont include $4$ cards in the same colour
My workings to get the answer above are:
$\dbinom{12}{4,8}$ Any selection of cards
$\dbinom74 \dbinom{5}{4}$ This is when there is either $4$ of red or $4$ of black.
Question
Would the answer $320$ be the correct answer?
There are only two colors, so you may first list all different size pairs first and then assign color for each of two stacks.
Consider stacks of size $3$ and $1$:
$(3, 1)$: ${7\choose 3}\cdot{5\choose 1} + {5\choose 3}\cdot{7\choose 1} = 35\cdot5+70=245.$
For this case: (3 red 1 black) + (3 black 1 red)
$(2, 2)$: ${7\choose 2}\cdot{5\choose 2} = 21\cdot 10 = 210$
So the total is $245+210=455.$