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A question goes like this: Given that there 2 questions out of 3 questions to pass an exam and follow are details of number of people who attended the questions correctly, what is the maximum number of people who could have qualified the exam.

1. 2 2 2

The answer here is 3. Consider ab bc ca. It can also be derived as 6/2 = (total number who cleared / number of questions which are required to clear).

If I give another sample data

2. 245 272 238

Even here I can say floor(755/2) = 377 are the maximum number of people who have passed the exam.

But consider this data.

3. 100 10 1

Now, the average formula would not work. By reasoning, I can say that 11 people could have passed the exam.

But mathematically, how do I derive this as in for any M tests (and we are given the number of people have cleared those M tests indivually) and minimum requirement is N tests, how we can find the maximum number of people who could have cleared the exam.

I hope, I explained it properly.

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Maybe it is just coincidence, but for (1) and (2), the maximum number is given by $S(\textbf{S},2)$ (the number of partitions of the sum into $2$ parts, where $\textbf{S}$ denotes the sum. See this: en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind –  PEV Jan 8 '11 at 19:10
    
Perhaps for "symmetric" sets of data, one can use the Stirling numbers of the second kind. For heavily skewed distributions you might have to use ad hoc methods. –  PEV Jan 8 '11 at 19:13
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1 Answer

For $M=3$ and $N=2$, this works:

floor $((\min(a,b+c) + \min(b,c+a) + \min(c,a+b))/2)$

For general $M,N$, I expect the answer to be much more complicated.

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