# Move elements in a grid (Combinatorics)

Here's an interesting and fairly simple problem I encountered a couple of weeks ago.

There is a grid with 11 rows and 11 columns with a ball in every cell. Move every ball to an adjacent cell (up, down, left or right - diagonals are not allowed). Show that no matter how you move the balls you will always end up with at least one cell with more than one balls in it.

• Hello, welcome to Math Stack Exchange! What do you want that we do? As you say that it is fairly simple, I assume you have solved it yourself, but then is our job kind of done... – wythagoras May 11 '15 at 17:11

Color the grid like a chessboard with alternating white and black squares. Let white be the color of the corners, so there are $61$ white squares and $60$ black squares. Notice that applying the move means balls that were on white squares are now on black squares, so since there were $61$ balls originally on white squares, but only $60$ black squares, by the pigeonhole principle after the move some (black) square must have more than $1$ ball.