Ways to Select Balls from Four Distinct Boxes

To find the total number of ways to select $10$ balls from $4$ distinct boxes such that at least one red ball and one blue ball are chosen from each box, let us analyze the possible selections from a single box.

Step 1: Ways to choose $k$ balls from a single box

Each box contains $3$ red balls and $2$ blue balls (all distinct).

Let $r$ be the number of red balls and $b$ be the number of blue balls chosen from a box. We are given $1 \le r \le 3$ and $1 \le b \le 2$. The total number of balls chosen from one box is $k = r + b$.

Let $W(k)$ be the number of ways to select $k$ balls from a single box satisfying the condition:

1. For $k = 2$:

   • $r = 1, b = 1$: $W(2) = \binom{3}{1} \times \binom{2}{1} = 3 \times 2 = 6$

2. For $k = 3$:

   • $r = 1, b = 2$: $\binom{3}{1} \times \binom{2}{2} = 3 \times 1 = 3$
   • $r = 2, b = 1$: $\binom{3}{2} \times \binom{2}{1} = 3 \times 2 = 6$
   • $W(3) = 3 + 6 = 9$

3. For $k = 4$:

   • $r = 2, b = 2$: $\binom{3}{2} \times \binom{2}{2} = 3 \times 1 = 3$
   • $r = 3, b = 1$: $\binom{3}{3} \times \binom{2}{1} = 1 \times 2 = 2$
   • $W(4) = 3 + 2 = 5$

4. For $k = 5$:

   • $r = 3, b = 2$: $\binom{3}{3} \times \binom{2}{2} = 1 \times 1 = 1$
   • $W(5) = 1$

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Step 2: Distribution of 10 balls among 4 boxes

Let $k_1, k_2, k_3, k_4$ be the number of balls selected from Box 1, Box 2, Box 3, and Box 4, respectively. We must have: $k_1 + k_2 + k_3 + k_4 = 10 \quad \text{where } 2 \le k_i \le 5 \text{ for each } i \in \{1, 2, 3, 4\}$

Let $k_i = 2 + x_i$, where $x_i \ge 0$. Substituting this gives: $x_1 + x_2 + x_3 + x_4 = 10 - 8 = 2$

The possible integer distributions of $(x_1, x_2, x_3, x_4)$ are:

Case 1: One $x_i = 2$ and three $x_i = 0$

This corresponds to selecting $(4, 2, 2, 2)$ balls from the four boxes in some order.

• The number of ways to assign these counts to the 4 distinct boxes is: $\frac{4!}{1! \, 3!} = 4$
• For each assignment, the number of ways to choose the balls is: $W(4) \times W(2)^3 = 5 \times 6^3 = 5 \times 216 = 1080$
• Total ways for Case 1: $4 \times 1080 = 4320$

Case 2: Two $x_i = 1$ and two $x_i = 0$

This corresponds to selecting $(3, 3, 2, 2)$ balls from the four boxes in some order.

• The number of ways to assign these counts to the 4 distinct boxes is: $\frac{4!}{2! \, 2!} = 6$
• For each assignment, the number of ways to choose the balls is: $W(3)^2 \times W(2)^2 = 9^2 \times 6^2 = 81 \times 36 = 2916$
• Total ways for Case 2: $6 \times 2916 = 17496$

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Step 3: Total Number of Ways

$\text{Total Ways} = 4320 + 17496 = 21816$

Thus, the correct option is (A).