A box contains 12 balls out of which 7 are red and rest are green. In how many ways can a random sample of 8 balls be drawn from the bag so that at the most 3 red balls are included in the sample and no sample has all the 8 balls of the same colour?

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7C3*5C5 =( 7!/3!*(7-3)! ) / 1 = 7*6*5*4! /3!*4! = 7*5 =35
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7C3*5C5=35
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Expert Mathematics teacher and tutor

It is not possible for all 8 balls to be of same color. Max 3 red and rest free = (7C3×5C5)!+ Max of 2 red balls=(7C2×5C6) not possible. Ans 7C3×5c5 random samples.
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35
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Math Magician

There are total 12 balls out of which 7 are red and 5 are green. This statement itself says the possibility of all 8 to be of same colour is nil. And to have at most 3 red, we require at least 5 green but we have only 5 green, this leave us with only one combination that is with 3 red and 5 green. No....
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There are total 12 balls out of which 7 are red and 5 are green. This statement itself says the possibility of all 8 to be of same colour is nil. And to have at most 3 red, we require at least 5 green but we have only 5 green, this leave us with only one combination that is with 3 red and 5 green. No. of samples= (7!/3!4!)*(5!/5!1!) =35 read less
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1 way
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Perfectionist

This question is too trivial. Answer: (7C3)x(5C5)=21. Replace 'atmost' with 'atleast' in the question. Then the solution will be (7C3)x(5C5)+(7C4)x(5C4)+ (7C5)x(5C3)+(7C6)x(5C2)+(7C7)x(5C1).
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Experienced Mathematics com ECE Educator

7C3*5C5= 35*1=35. there are 35 possible ways
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Tutor

7C3*5C5= 35,since 6 balls cannot be choosen form 5 balls and even more hence 35 is correct answer
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