1985 Putnam A1 Solution

For any element $x \in \{1,\ldots,10\}$, in order to satisfy the constraints, $x$ must belong to either exactly 1 or exactly 2 of the sets $A_1,A_2,A_3$. This gives you 6 possibilities per element (why?), for a total of $6^{10}$ possibilities.

Don't bother with the matrix; this is just a fancy way of stating the same argument.

For example, suppose the three sets are \begin{align} A_1 & = \{1,5,6,8\} \\ A_2 & = \{1,2,3,4,10\} \\ A_3 & = \{2,4,5,7,9,10\} \end{align} (Note that $A_1\cup A_2\cup A_3=\{1,2,3,4,5,6,7,8,9,10\}$ and $A_1\cap A_2\cap A_3=\varnothing$.)

Then the $10\times3$ matrix is $$ \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} $$ The first column of the matrix corresponds to the set $A_1$. It has a $1$ in the $1$st, $5$th, $6$th, and $8$th rows because the members of $A_1$ are $1,5,6,8$. Similarly the second and third columns correspond respectively to $A_2$ and $A_3$.