Counting integers starting from 2

Case-I: If zero is on the 2^nd place

i.e., $2\;0\;2\;\underbrace{\uparrow}_{5} \;\Rightarrow\; 5\text{ cases}$

or $2\;0\;\underbrace{\uparrow\;\uparrow}_{4,\;6} \;\Rightarrow\; 24\text{ cases}$

(Numbers except 0 or 2 in the 3^rd place)

Case-II: If a non-zero number is on the 2^nd place

i.e., $2\;\underbrace{\uparrow}_{5}\; \underbrace{\uparrow}_{6}\; \underbrace{\uparrow}_{6} \;=\;180\text{ cases}$

Counting integers starting from 3

$3\;\underbrace{\uparrow}_{6}\; \underbrace{\uparrow}_{6}\; \underbrace{\uparrow}_{6} \;=\;216\text{ cases}$

Counting integers starting from 4

Case-I: If 0, 2 or 3 is on the 2^nd place

i.e., $4\;\underbrace{\uparrow}_{3}\; \underbrace{\uparrow}_{6}\; \underbrace{\uparrow}_{6} \;=\;108\text{ cases}$

Case-II: If 4 is on the 2^nd place

i.e., $4\;4\; \underbrace{\uparrow}_{6}\; \underbrace{\uparrow}_{6} \;=\;36\text{ cases}$

$\therefore$ Total $=5+24+180+216+108+36$

$\boxed{569\text{ numbers}}$