TITLE: Unique matrix of zeros and ones
QUESTION [6 upvotes]: We have a matrix of zeros and ones $a_{ij}$ with 16 rows and 30 columns. ($a_{ij}=\{0$ or $1\}$). 

For any rows $i=m$ : $\sum\limits_{j=1}^{30}a_{mj} \leq 15$. 
For any column $j=n$ : $\sum\limits_{i=1}^{16}a_{in} \geq 8$. 
Any 2 rows have exactly k ones ("1") in the same column. 


k = ? Using brute force I find $k=7$. I'm looking for some nice and simple mathematical solution how to prove this.

REPLY [4 votes]: I've been thinking about your question about a combinatorial way to construct the matrix.  Below is a diagram to help me explain what I did, representing the 1's as boxes and the 0's as empty space.

You know that you're going to end up with exactly fifteen 1's in each row and exactly eight 1's in each column.  It doesn't matter what order the rows and columns are in, so without loss of generality we can gather the 1's in the first row on the left (i.e. $a_{1j}=1$ if and only if $j \leq 15$).  You know you want the second row to share exactly 7 1's with the first, so again without loss of generality we can let $a_{2j}=1$ for $j=1,\ldots,7$.  We must have 0's for the remaining spaces under the first row's 1's, so the other eight 1's go just to the right.
Now, it becomes helpful to notice the symmetry of the matrix we're trying to make.  We're also making a matrix with exactly fifteen 0's in each row and eight 0's in each column, and by gathering the 1's on the left, we're gathering the 0's on the right.  Thus any row we make must be able to be symmetric in the "negative space," as an artist would say - in other words, we could reverse the order of the row, switch the 1's and 0's, and get the same thing back.
I continued in this manner, trying to put as many 1's on the left as I could.   Because we need 15 1's in a row, I could not bring 5 on the left while maintaining symmetry.  Thus I found two rows with 3 on the left.  Knowing that the top half of the columns must add to 4 (again by symmetry), rows 5 through 8 were then forced to be filled in as illustrated.  After row 8, I simply reflected the top half of the columns onto the bottom half, then inverted the 1's and 0's in the bottom half of every other column to enforce that the arrangements were unique.