Consider 4 boxes, where each box contains 3 red balls and 3 blue balls. Assume that all 24 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

$\boxed{\begin{array}{c}3R\\ 3B\end{array}}\boxed{\begin{array}{c}3R\\ 3B\end{array}}\boxed{\begin{array}{c}3R\\ 3B\end{array}}\boxed{\begin{array}{c}3R\\ 3B\end{array}}$

Case : 1) 4, 2, 2, 2
Case : 2) 3, 3, 2, 2
1)  $\left(4C_1\right){(}^3{C}_3{.}^3{C}_1{+}^3{C_2}^3{C}_2{+}^3{C_1}^3{C}_3){(}^3{C}_1{.}^3{C}_1){(}^3{C}_1{.}^3{C}_1){(}^3{C}_1{.}^3{C}_1)$
 $\left(4\right)(3+9+3)\left(9\right)\left(9\right)\left(9\right)=43,740$
2)  $\left(4C_2\right){(}^3{C}_1{.}^3{C}_2{+}^3{C_2}^3{C}_1){(}^3{C}_1{.}^3{C}_2{+}^3{C}_2{.}^3{C}_1){(}^3{C}_1{.}^3{C}_1){(}^3{C}_1{.}^3{C}_1)$
 $\left(6\right)(9+9)(9+9)\left(9\right)\left(9\right)=15,7,464$
No. of ways = 43, 740 + 1, 57, 464 = 2, 01, 204