JEE Main 2026 — Mathematics Algebra

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

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

The possible combinations for $(B, Y, R)$(B,Y,R) are:

Case 1: $(2, 2, 4)$(2,2,4)

Case 2: $(2, 3, 3)$(2,3,3)

Case 3: $(2, 4, 2)$(2,4,2)

Case 4: $(3, 2, 3)$(3,2,3)

Case 5: $(3, 3, 2)$(3,3,2)

Case 6: $(4, 2, 2)$(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$5C2​×6C2​×4C4​=10×15×1=150

Case 2: $^{5}C_{2} \times ^{6}C_{3} \times ^{4}C_{3} = 10 \times 20 \times 4 = 800$5C2​×6C3​×4C3​=10×20×4=800

Case 3: $^{5}C_{2} \times ^{6}C_{4} \times ^{4}C_{2} = 10 \times 15 \times 6 = 900$5C2​×6C4​×4C2​=10×15×6=900

Case 4: $^{5}C_{3} \times ^{6}C_{2} \times ^{4}C_{3} = 10 \times 15 \times 4 = 600$5C3​×6C2​×4C3​=10×15×4=600

Case 5: $^{5}C_{3} \times ^{6}C_{3} \times ^{4}C_{2} = 10 \times 20 \times 6 = 1200$5C3​×6C3​×4C2​=10×20×6=1200

Case 6: $^{5}C_{4} \times ^{6}C_{2} \times ^{4}C_{2} = 5 \times 15 \times 6 = 450$5C4​×6C2​×4C2​=5×15×6=450

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

Answer: $4100$4100