The answer sheets of 5 engineering students can be checked by any one?
First. let's see what are the total number of ways in which the 5 answer sheets can be checked. For getting one particular answer sheet checked, we have 9 choices(professors). And we want to get all of them checked. Also, each sheet is independent of other, as in, one sheet being checked by a professor does not limit the choices for other sheets.This no. should be T = 9 * 9 * 9 * 9 * 9 = 9^{5} . To get a better insight into this, let's pick up one of the combination in this set and try to see what it means,P1, P3, P5, P1, P7 is a way in which first sheet is checked by Professor 1, second by Professor 3, third by Professor 5, fourth by Professor 1, and fifth by Professor 7. We could change the professor for first sheet to Professor 8, and that would be another combination. In fact we could change the professor for any of the sheets and that would be a different combination all together. Now, we want to calculate the no. of ways in which all five answer sheets are checked by exactly 2 professors. First let's see the no. of ways in which the answer sheets could be checked if those were to be checked by exactly two professors A and B. We don't know who professor A and B are at the moment, all we know is A and B are two different professors from the set {P1, P2, ... , P9}.Now, if we want to get the 5 sheets checked by exactly A and B, it seems to be the no. of onto functions we can have from set P ({S1, S2, S3, S4, S5}) to set Q({A, B}). right?Because, for each sheet there has to be a professor and exactly one professor who should check it, and each professor (A and B) must check at least one sheet, only then we can say that exactly two of them checked the sheets and not just one of them. Now, if we want to get no. of onto functions from set P to set Q, we will first get the total no. of functions from P to Q, and then remove the ones which are not onto.First, what is the total no. of functions from P to Q. This is similar to previous part, we want to have a value(y) for each of the sheets (x). There are two choices for each of the sheets(A/B) and there are five sheets, so the total no. is 2^{5} . But this involves those functions as well in which all the sheets were checked by exactly one professor (i.e. either A or B). The no. of ways in which exactly one professor checked all the sheets is 2. Either all the sheets get checked by A or all of them get checked by B. So, total no. of ways in which all the sheets are checked by exactly two professors is 2^{5} - 2. But, this was when the professors who were destined to check the sheets were limited to A and B. We should chose two professors who will play the part of A and B. This can be done by choosing any two professors from the set {P1, P2, .. P9}. It doesn't matter who is assigned label A and who is assigned label B, because we never distinguished between the two labels. Since, for each pair of professors from the set {P1, P2, P3, ... P9}, same no. of ways will be there to get an onto function, the total no. of ways in which exactly two professors check all the five sheets is N= {9 \choose 2} (2^{5} -2). Hence, the final probability comes out to be N/T. Of course this makes sense only if a sheet is equally likely to be check by any professor and the same is true for all the sheets. :)P.S: Do comment if you don't follow the logic anywhere or if you feel there is a flaw in the approach.