If eight persons are to address a meeting then the number of ways in which a specified speaker is to speak before another specified speaker, is

Let A, B be the corresponding speakers. Without any restriction the eight persons can be arranged among themselves in 8! ways, but the number of ways in which A speaks before B and the number of ways in which B speaks before A make up 8!. Also, number of ways in which A speaks before B is exactly same as the number of ways in which B speaks before A.

$\therefore$ the required number of ways $=\frac{1}{2}(8!)=20160$.