A box contains 5 blue, 6 yellow and 4 red balls. The number of...

Let $B$, $Y$, and $R$ be the number of blue, yellow, and red balls drawn respectively.

We are given $B + Y + R = 8$ with $B \ge 2$, $Y \ge 2$, and $R \ge 2$.

The possible combinations for $(B, Y, R)$ are:
Case 1: $(2, 2, 4)$
Case 2: $(2, 3, 3)$
Case 3: $(2, 4, 2)$
Case 4: $(3, 2, 3)$
Case 5: $(3, 3, 2)$
Case 6: $(4, 2, 2)$

The number of ways for each case is calculated as follows:
Case 1: $^{5}C_{2} \times ^{6}C_{2} \times ^{4}C_{4} = 10 \times 15 \times 1 = 150$
Case 2: $^{5}C_{2} \times ^{6}C_{3} \times ^{4}C_{3} = 10 \times 20 \times 4 = 800$
Case 3: $^{5}C_{2} \times ^{6}C_{4} \times ^{4}C_{2} = 10 \times 15 \times 6 = 900$
Case 4: $^{5}C_{3} \times ^{6}C_{2} \times ^{4}C_{3} = 10 \times 15 \times 4 = 600$
Case 5: $^{5}C_{3} \times ^{6}C_{3} \times ^{4}C_{2} = 10 \times 20 \times 6 = 1200$
Case 6: $^{5}C_{4} \times ^{6}C_{2} \times ^{4}C_{2} = 5 \times 15 \times 6 = 450$

Total number of ways $= 150 + 800 + 900 + 600 + 1200 + 450 = 4100$

Answer: $4100$