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# The no. of ways in which 10 students $A_1,A_2,.....A_{10}$ can be ranked so that $A_1$ is above $A_2$ is ?

$\begin{array}{1 1} (10)! \\ \frac{10 !}{2} \\ 9! \\ \frac{9!}{2} \end{array}$

The no of ways in which 10 students are ranked = $10!\: ways$
Out of these in half of the ways $A_1$ is above $A_2$ and
in remaining half $A_2$ is above $A_1$.
$\therefore$ The required no. of ways $= \large\frac{10!}{2}$