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Maths Question 9 – JEE-MAIN 2026

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A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is :

The problem requires drawing 8 balls with a minimum of two balls of each color from a given set of blue, yellow, and red balls.

Step 1: Define variables and constraints✦ Active

Let b,y,r be the number of blue, yellow, and red balls drawn, respectively. The total number of balls to be drawn is 8, so b+y+r=8. The available balls are 5 blue, 6 yellow, and 4 red. The condition is to draw at least two balls of each color. This implies the following constraints:

2≤b≤5 2≤y≤6 2≤r≤4
Step 2: List all valid combinations of (b,y,r)○ Expand

Since we must have at least 2 balls of each color, 2+2+2=6 balls are already accounted for. We need to distribute the remaining 8−6=2 balls among the three colors, respecting the upper limits. The possible combinations (b,y,r) are:

(4,2,2)(2,4,2)(2,2,4) (3,3,2)(3,2,3)(2,3,3)
💡 Teacher's Secret Hint

Ensure all combinations satisfy both the sum of 8 and the individual upper limits for each color.

Step 3: Calculate ways for each combination and sum them up○ Expand

Using the combination formula C(n,k)=n!k!(n−k)!:

C(5,4)C(6,2)C(4,2)=5×15×6=450 C(5,2)C(6,4)C(4,2)=10×15×6=900 C(5,2)C(6,2)C(4,4)=10×15×1=150 C(5,3)C(6,3)C(4,2)=10×20×6=1200 C(5,3)C(6,2)C(4,3)=10×15×4=600 C(5,2)C(6,3)C(4,3)=10×20×4=800

The total number of ways is the sum of these possibilities:

450+900+150+1200+600+800=4100
💡 Teacher's Secret Hint

Double-check each combination calculation to avoid arithmetic errors.

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