The number of $4$-digit integers in the closed interval $\left[2022,4482\right]$ formed by using the digits $0,2,3,4,6,7$ is ______.

Given the number to be formed by using the digits $0,2,3,4,6,7$ 
Now assuming counting integers starting from $2$
Case-I: if zero on $2^{\mathrm{nd}}$ place
(i) $2 0 2 \_$, now here at unit place $5$ number can take place except zero {as $2020<2022$}, so $5$ possible way.
(ii) $2 0 \_ \_$, now here at ten's place $4$ number $\left\{3,4,6,7\right\}$ can take place  and at unit place all $6$ number can take place so total $4\times 6=24$ possible way
Case-II: If non-zero number on $2^{\mathrm{nd}}$ place
i.e $2\underset{5}{ \underbrace{ \_}} \underset{6}{\underbrace{\_}}\underset{6}{ \underbrace{ \_}}$, here at hundredth place $5$ possibility will be there and at unit and ten's place $6 \& 6$ possibility will be there, so in total $5\times 6\times 6=180$ possible cases,
Case -III when counting integers starting from $3$
$\begin{array}{cccc}3& \underset{6}{\underbrace{\_}}& \underset{6}{\underbrace{\_}}& \underset{6}{\underbrace{\_}}\end{array}$ here total $6\times 6\times 6=216$ possible cases
Case -IV when counting integers starting from $4$
(i) If $0,2$ or $3$ on $2^{\mathrm{nd}}$ place
i.e., $\begin{array}{cccc}4& \underset{3}{\underbrace{\_}}& \underset{6}{\underbrace{\_}}& \underset{6}{\underbrace{\_}}\end{array}$ here total $3\times 6\times 6=108$ possible cases
(ii) If $4$ on $2^{\mathrm{nd}}$ place
i.e., $\begin{array}{cccc}3& 4& \underset{6}{\underbrace{\_}}& \underset{6}{\underbrace{\_}}\end{array}$ here total $6\times 6=36$ possible cases
So, total possible cases will be $5+24+180+216+108+36=569$ numbers