let number the students as 1,2,3, 3n then a particular arrangement of leaving would be like 23,3,7,18,4,15,29,1...... i assume here all such leaving arrangement has equal chance or probability to occur so total no of such arrangement is (3n)! now consider what happens for the events induced by the constraint \[ |N_{i}(t)-N_{j}(t)| < 2, i\neq j, i, j = 1, 2,\dots . \] let wlog first leaving guy is from row1 ,then the second one have to be from the other two row say row 3, then third one must be from row 2 so its like sequence of three blocks where in each block contains one from each row no of such arrangements is $ \ 6^n(n!)^3 $ so the answer is $ \frac{6^n(n!)^3}{3n!} $