Number of Ways to Draw Balls with Minimum Constraints

To find the total number of ways of drawing $8$ balls containing at least two balls of each colour, let $B$, $Y$, and $R$ represent the number of blue, yellow, and red balls drawn, respectively.

We are given:

• Total blue balls available $= 5 \implies B \le 5$
• Total yellow balls available $= 6 \implies Y \le 6$
• Total red balls available $= 4 \implies R \le 4$

The conditions for the draw are: $B + Y + R = 8$ $B \ge 2, \quad Y \ge 2, \quad R \ge 2$

We can list all possible valid combinations of $(B, Y, R)$ that satisfy these conditions and calculate the number of ways to choose them using combinations $\binom{n}{r}$:

1. Case 1: $(B, Y, R) = (4, 2, 2)$ $\text{Ways} = \binom{5}{4} \times \binom{6}{2} \times \binom{4}{2} = 5 \times 15 \times 6 = 450$

2. Case 2: $(B, Y, R) = (2, 4, 2)$ $\text{Ways} = \binom{5}{2} \times \binom{6}{4} \times \binom{4}{2} = 10 \times 15 \times 6 = 900$

3. Case 3: $(B, Y, R) = (2, 2, 4)$ $\text{Ways} = \binom{5}{2} \times \binom{6}{2} \times \binom{4}{4} = 10 \times 15 \times 1 = 150$

4. Case 4: $(B, Y, R) = (3, 3, 2)$ $\text{Ways} = \binom{5}{3} \times \binom{6}{3} \times \binom{4}{2} = 10 \times 20 \times 6 = 1200$

5. Case 5: $(B, Y, R) = (3, 2, 3)$ $\text{Ways} = \binom{5}{3} \times \binom{6}{2} \times \binom{4}{3} = 10 \times 15 \times 4 = 600$

6. Case 6: $(B, Y, R) = (2, 3, 3)$ $\text{Ways} = \binom{5}{2} \times \binom{6}{3} \times \binom{4}{3} = 10 \times 20 \times 4 = 800$

Summing up the ways from all valid cases gives the total number of ways: $\text{Total ways} = 450 + 900 + 150 + 1200 + 600 + 800 = 4100$

Thus, the correct option is A.