Question 1062843: An instructor gives an exam with fourteen questions. Students are allowed to choose any ten to answer.
Suppose six questions require proof and eight do not.
1-How many groups of ten questions contain at least one that requires proof?
2-How many groups of ten questions contain at most three that require proof?
My answers
for 1: C(6,2)*C(8,8)+C(6,3)*C(8,7)+C(6,4)*C(8,6)+C(6,5)*C(8,5)+C(6,6)*C(8,4)
for 2: C(6,2)*C(8,8)+C(6,3)*C(8,7)
i am not sure if they are correct
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! --
For (1): No matter how you select 10 questions from the 14, one or more of the 10 will have to require proof (only 8 do not require proof so there's always at least 2 that require proof). Thus I think the answer is "all possible selections" and that is C(14,10).
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For (2): I'd reason as follows: C(14,10) have at least 2 questions that require proof. The maximum we care about is 3 questions requiring proof, so from the C(14,10) you'll need to subtract the number of groups with exactly {4 or 5 or 6} requiring proof.
C(6,4)*C(8,6) = # with 4 requiring proof
C(6,5)*C(8,5) = # with 5 requiring proof
C(6,6)*C(8,4) = # with 6 requiring proof
So I think (2) is: C(14,10) - C(6,4)*C(8,6) - C(6,5)*C(8,5) - C(6,6)*C(8,4)
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For #1, our answers agree C(14,10)=1001 and so does your answer. Although we took different approaches, we both arrived at the same result. It is likely you are correct.
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For #2, our answers agree as well, value is 175 for both.
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I can't explain why, but I see my way of looking at (2) as a "take away" problem while your solution is more of a "constructive" or additive solution. Sometimes that happens I guess. Anyway, good work!
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