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From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen

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asked Jan 5, 2018 in Mathematics by Golu (37,045 points) ● 19 ● 169 ● 595

From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking at least one green and one blue ball?

1 Answer

+1 vote
answered Jan 5, 2018 by Rohit Singh (61,782 points) ● 36 ● 142 ● 452
selected Jan 5, 2018 by Golu
 
Best answer

Total combination of balls = 212
Combination of balls with no green balls = 27
Combination of balls with no blue balls = 28
Combination of balls with no blue balls  and no green balls = 23

Required combination = 212 - 27 - 28 + 23 = 3720

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