Consider 4 boxes, where each box contains 3 red balls and 2 blue...

Let the four boxes be Box-1, Box-2, Box-3 and Box-4. Each of the box has $3$ red balls and $2$ blue balls.

$10$ balls can be chosen in:

Case I - When exactly one box gives four balls

Case II - When exactly two boxes gives three balls

Required number of ways= Number of ways in Case I $+$ Number of ways in Case II.

$= C1 4 \left[C1 2 + C1 3\right]{\left(C1 3 C1 2\right)}^{3} + C2 4 {\left[C2 3 C1 2 + C1 3 C2 2\right]}^{2}{\left[C1 3 C1 2\right]}^{2}$

$= 4\left(5\right){\left(6\right)}^{3} + 6{\left(3 \times 2 + 3\right)}^{2}{\left(6\right)}^{2}$

$= 4320 + 17496 = 21816$