The number of 4 -digit integers in the closed interval [2022, 4482]...

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 ${2}^{nd}$ place

(i) 202_, now here at unit place $5$ number can take place except zero {as $2020 < 2022$}, so $5$ possible way.

(ii) 20__, 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</p><p>Case-II: If non-zero number on${2}^{nd}$place</p><p>i.e$2\mathop{\mathop{_}_{⏟}}_{5}\mathop{\mathop{_}_{⏟}}_{6}\mathop{\mathop{_}_{⏟}}_{6}$, here at hundredth place$5$possibility will be there and at unit and ten's place$6 \text{ and } 6$possibility will be there, so in total$5 \times 6 \times 6 = 180$possible cases,</p><p>Case -III when counting integers starting from$3$</p><p>$\begin{cases}3 & \mathop{\mathop{_}_{⏟}}_{6} & \mathop{\mathop{_}_{⏟}}_{6} & \mathop{\mathop{_}_{⏟}}_{6}\end{cases}$here total$6 \times 6 \times 6 = 216$possible cases</p><p>Case -IV when counting integers starting from$4$</p><p>(i)$ If 0, 2 or 3$on${2}^{nd}$place</p><p>i.e.,$\begin{cases}4 & \mathop{\mathop{_}_{⏟}}_{3} & \mathop{\mathop{_}_{⏟}}_{6} & \mathop{\mathop{_}_{⏟}}_{6}\end{cases}$here total$3 \times 6 \times 6 = 108$possible cases</p><p>(ii)$ If 4$on${2}^{nd}$place</p><p>i.e.,$\begin{cases}3 & 4 & \mathop{\mathop{_}_{⏟}}_{6} & \mathop{\mathop{_}_{⏟}}_{6}\end{cases}$here total$6 \times 6 = 36$possible cases</p><p>So, total possible cases will be$5 + 24 + 180 + 216 + 108 + 36 = 569 numbers