You want to put several balls on $8 \times 8$ tiles, such that all $16$ ball arrangements on its rows and columns are different. What is the minimum number of balls to be put?

Two arrangements of balls are different if there exists a ball on some $i$-th tile of the first one but no ball on $i$-th tile of the second one. (For row, the $i$-th tile means the $i$-th column; for column, the $i$-th tile means the $i$-th row.)

For example, we need at least $4$ balls for $3 \times 3$ tiles and the configuration as the following:

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Bonus: How about $N \times N$?


A solution in:



Borrowing from @Jaap:



  • $\begingroup$ Beat me to it. And a better solution is clearly not possible. $\endgroup$ – Daniel Mathias Feb 10 at 15:29

I think this might be an optimal solution, but I’m not sure. (Edit: It's not optimal - one fewer balls is possible)


This generalises in an obvious way for even $N$ to give a solution with $3N/2$ balls. For odd $N$ you can do almost the same, but with one fewer row an column with two balls, giving a total of $(3N-1)/2$ balls.

  • $\begingroup$ nicely done! Your answer inspired me to solve mine. You were so close! The general formula was only slightly off. Thanks @Jaap $\endgroup$ – Krad Cigol Feb 11 at 14:09

Piggybacking off @JonMarkPerry:

The answer is 11.


We have a total of 2N rows and columns, with each having a unique permutation of balls.
We get N c 1 ways to place 1 ball.
We get N c 2 ways with 2,
And so on.
Now, we need to find the minimal number of balls to give each row a unique permutation.
Note that N c 1 = N.
N c 2 = N(N-1)
And so on.

Adding the two, we get

Which is clearly greater than 2N, assuming that $N>2$

So the requirement is:

0 balls for 1 row/column.
$N*1$ balls for next N row/columns.
Remaining N require $2(2N-N-1)=2N-2$
This adds up to a total minimum of $3N-2$ We need to divide by two, as we counted each ball twice, once in a row and once in a column.


The general formula is

  • $\begingroup$ what if N is odd? $\endgroup$ – JonMark Perry Feb 10 at 16:03
  • $\begingroup$ We round it down, @JonMarkPerry. $\endgroup$ – Krad Cigol Feb 10 at 16:12
  • $\begingroup$ By the way, I think I missed the one row without balls. And a few other errors. Will fix it later. It adds a good amount of error to the equation. Sorry, @JonMarkPerry! $\endgroup$ – Krad Cigol Feb 10 at 16:13
  • $\begingroup$ how does your general formula give the answer as 11 for N=8? $\endgroup$ – JonMark Perry Feb 10 at 21:45
  • 2
    $\begingroup$ this answer provides the lower bound, however how to construct the arrangements for such lower bound? :) $\endgroup$ – athin Feb 11 at 23:00

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