Three natural numbers are taken at random from the set A= {x : 1 ≤ x ≤ 100, x ∈ N. then the probability of that AM of the number taken is 25

$\frac{{ }^{77} C_2}{{ }^{100} C_3}$ $\frac{{ }^{25} C_2}{{ }^{100} C_3}$ $\frac{{ }^{74} C_2}{{ }^{100} C_3}$ none of these

$\frac{{ }^{74} C_2}{{ }^{100} C_3}$

$x_1+x_2+x_3=75$ coefficient $x^{75}$ in $\left(x^1+x^2+....+x^{100}\right)^3={ }^{74} C_2$ Hence probability $=\frac{{ }^{74} C_2}{{ }^{100} C_3}$.