Combinatoric problem from British Math Olympiad, 2009/2010 This is question 1 from the 2009/10 issue of the BMO, and it reads as follows:
There are   $2010^{2010}$    children at a mathematics camp. Each has at most 3 friends at the camp, and if A is friend with B, B is friend with A. The camp leader would like to line up the children such that there are 2010 children between any pair of friends. Is this always possible?
I was wondering is anyone had any clues/hints on how to solve this?
 A: I will share my thoughts are about this question. So As it is given if A if friends with B then it is vice versa also true ,So  Let's start with some child. And then you will have to choose 3 friends for him out this  $2010^{2010} $. So Now after deciding for him you will have these three friends of your first person and each of them need two more friends , Now You provide them with two more friends and continue this task( supposing it is a very big number so this task will move up infinitely. One thing to observe is that it forms a tree Like Structure like this

C-> Is the initial Child,
F-> Friend OK
After covering up N levels of this tree like structure we might have encountered
$3 * 2^N -2 $  children.
And For these many children we have encountered We have to line each pair with at least $2010$ children between each of them So these many Children might fit into $2011*N$ size block to satisy the above property .
All you have to do now is check if this inequality gives desired results or not
$ 2011*N >= 3 * 2^N -2 $
My result says that it will happen to give false results for Large Values of N and it will be impossible to fit children hence
