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?
(3i Infotech Question)
Asked by anshu
arun kumar   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
3 years ago
  
2 years ago

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