1985 Putnam A1 Solution 
I dont see what they mean by bijection of triples of subsets of $\{1, \ldots, 10\}$ and the $10\times3$ matrix with $0, 1$ entries? 
How is that created? 
 A: 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.
A: 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 $A_3$.
