Three natural numbers are taken at random from the set $\mathrm{A}=\left\{\mathrm{x}|1\le \mathrm{x}\le 100,\mathrm{x}\in \mathrm{N}|\right\}$. The probability that the AM of the numbers taken is 25, is:

$\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3}=25\Rightarrow {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3=75$
Favorable ways = coefficient  of x⁷⁵ in ${\left(\mathrm{x}+{\mathrm{x}}^2+\dots .+{\mathrm{x}}^{100}\right)}^3$
 = coefficient of x⁷² in ${\left[\frac{1-{\mathrm{x}}^{100}}{1-\mathrm{x}}\right]}^3$
= coefficient of x⁷² in (1-x)^-3        
= ${ }^{74}{\mathrm{C}}_{72}$
Total number of ways of selecting 3 numbers from $1.2100{=}^{100}{\mathrm{C}}_3{=}^{100}{\mathrm{C}}_{97}\text{. }$
$\therefore$Required probability = $\frac{{ }^{74}{\mathrm{C}}_{72}}{{{100}_{\mathrm{C}}}_{{ }_{97}}}$