Indian Institute of Foreign Trade 2012 Solved Paper

Each of the 5 papers can be checked by any of the 9 professors and thus the total number of way $=9^5$
Selecting exactly 2 of the 9 professors can be done in ${}_{}^9{\mathit{C}}_2^{}$ way = 36
They can correct the 5 papers in $2^5-2$(all the 5 papers checked by the same professor) = 30 ways.
Thus, the total number of ways $36\times 30$
Hence, the required probability $=\frac{36\times 30}{9^5}=\frac{40}{2187}$
Hence, option B is the correct answer.