We number the columns of an n×n-board from 1 to n. In each cell, we place a number. This is done in such a way that each row precisely contains the numbers 1 to n (in some order), and also each column contains the numbers 1 to n (in some order). Next, each cell that contains a number greater than the cell's column number, is coloured grey. In the figure below you can see an example for the case n=3.[asy]
unitsize(0.6 cm);int i;fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.8));
fill(shift((1,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), gray(0.8));
fill(shift((0,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), gray(0.8));for (i = 0; i <= 3; ++i) {
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}label("1", (0.5,3.5));
label("2", (1.5,3.5));
label("3", (2.5,3.5));
label("3", (0.5,2.5));
label("1", (1.5,2.5));
label("2", (2.5,2.5));
label("1", (0.5,1.5));
label("2", (1.5,1.5));
label("3", (2.5,1.5));
label("2", (0.5,0.5));
label("3", (1.5,0.5));
label("1", (2.5,0.5));
[/asy](a) Suppose that n=5. Can the numbers be placed in such a way that each row contains the same number of grey cells?
(b) Suppose that n=10. Can the numbers be placed in such a way that each row contains the same number of grey cells? boardsquare gridColoringcombinatorics