Question 1062843
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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!