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Permutation and Combination
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?
Read Solution (Total 3)
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- combination of atleast 1 green ball: 5C1+5C2+5C3+5C4+5C5=31 ways
combination of atleast 1 blue ball:4C1+4C2+4C3+4C4=15 ways
combination of red balls:3C0+3C1+3C2+3C3=8 ways
total combination:31*15*8= 3720 - 11 years agoHelpfull: Yes(33) No(8)
- Total combination of balls = 2^12
Combination of balls with no green balls = 2^7
Combination of balls with no blue balls = 2^8
Combination of balls with no blue balls and no green balls = 2^3
Required combination = 2^12 - 2^7 - 2^8 + 2^3 = 3720 - 7 years agoHelpfull: Yes(0) No(0)
- combination of atleast 1 green ball: 5C1+5C2+5C3+5C4+5C5=31 ways
combination of atleast 1 blue ball:4C1+4C2+4C3+4C4=15 ways
combination of red balls:3C0+3C1+3C2+3C3=8 ways
total combination:31*15*8= 3720 - 5 years agoHelpfull: Yes(0) No(0)
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