²  For a fixed sample size \(N\), a \(2\times 2\) contingency table can be represented by nonnegative integer cell counts \((a,b,c,d)\) satisfying \(a+b+c+d=N\). The number of such tables is \(\binom {N+3}{3}\) (equivalently, \({}_4H_N\)), which is the same as the number of ways to choose three balls from a sequence of \(53\) balls to determine the number of balls between and beyond the three chosen balls as the desired partition of \(50\). For \(N=50\), this gives \({}_4H_{50}=\binom {53}{3}=23{,}426\) possible tables. We then excluded tables with two or more zero cells. The number of tables with exactly two zero cells is \(\binom {4}{2}(N-1)\): there are \(\binom {4}{2}=6\) ways to choose which two cells are zero, and the remaining two positive counts must sum to \(N\), which yields \(N-1\) possibilities. The number of tables with exactly three zero cells is 4 (all observations fall in a single cell). Thus, the number excluded is \(6(N-1)+4\), which equals \(298\) when \(N=50\), leaving \(23{,}426-298=23{,}128\) tables.