Consider a set $\{1,2,3,\mathrm{.....},100\}.$ The number of ways in which a number can be selected from the set so that it is of the form $x^y,$ (where $x,y\in N$ and $\ge 2$ ), is

$\text{Perfect square}=\left[\sqrt{100}\right]-1=9\left(\text{excluding one}\right)$

$\text{Perfect cubes}=\left[{100}^{1/3}\right]-1=3$

${\text{Perfect 4}}^{th}\text{\hspace{0.17em}powers}=\left[{100}^{1/4}\right]-1=2$

${\text{Perfect 5}}^{th}\text{\hspace{0.17em}powers}=\left[{100}^{1/5}\right]-1=1$

${\text{Perfect 6}}^{th}\text{\hspace{0.17em}powers}=\left[{100}^{1/6}\right]-1=1$

Now, perfect $4^{th}$ powers have already been counted in perfect squares and perfect 6th powers have been counted with perfect squares as well as with perfect cubes Hence the total ways = 9 + 3 + 1 – 1 = 12