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+1 vote
38.5k views
in Mathematics by (106k points)

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

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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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