n2. Under Hardy-Weinberg equilibrium, each of these ways is equally likely. Thus, the ratio between the likelihoods of observing exactly n12=k+2 heterozygotes and exactly n12=k heterozygotes, under Hardy-Weinberg equilibrium and fixed n1 and n2, is \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\begin{array}{@{}rcl@{}} & \left(\frac{(2^{k+2})(n!)}{(\frac{n_{1}-k}{2}-1)!(k+2)!(\frac{n_{2}-k}{2}-1)!} \middle/ \frac{(2^{k})(n!)}{\frac{n_{1}-k}{2}!k!\frac{n_{2}-k}{2}!} \right) \\ = & \frac{2^{k+2}}{2^{k}}\cdot \frac{n!}{n!}\cdot \frac{\frac{n_{1}-k}{2}!}{(\frac{n_{1}-k}{2}-1)!}\cdot \frac{k!}{(k+2)!}\cdot \frac{\frac{n_{2}-k}{2}!}{(\frac{n_{2}-k}{2}-1)!} \\ = & 4\cdot 1\cdot \frac{n_{1}-k}{2}\cdot \frac{1}{(k+1)(k+2)}\cdot \frac{n_{2}-k}{2} \\ = & \frac{(n_{1}-k)(n_{2}-k)}{(k+1)(k+2)}. \end{array} $$ \end{document}(2k+2)(n!)(n1−k2−1)!(k+2)!(n2−k2−1)!/(2k)(n!)n1−k2!k!n2−k2!=2k+22k·n!n!·n1−k2!(n1−k2−1)!·k!(k+2)!·n2−k2!(n2−k2−1)!=4·1·n1−k2·1(k+1)(k+2)·n2−k2=(n1−k)(n2−k)(k+1)(k+2).
