JEE Advanced Mathematics Question (2022) — Solution

Step-by-step solution

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)  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 3 , 4 , 6 , 7  can take place  and at unit place all  6  number can take place so total  4 × 6 = 24  possible way Case-II: If non-zero number on  2 nd  place i.e  2     _ ⏟ 5     _ ⏟ 6     _ ⏟ 6 , 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 × 6 × 6 = 180  possible cases, Case -III when counting integers starting from  3 3 _ ⏟ 6 _ ⏟ 6 _ ⏟ 6  here total  6 × 6 × 6 = 216  possible cases Case -IV when counting integers starting from  4 (i) If  0 , 2  or  3  on  2 nd  place i.e.,  4 _ ⏟ 3 _ ⏟ 6 _ ⏟ 6  here total  3 × 6 × 6 = 108  possible cases (ii) If  4  on  2 nd  place i.e.,  3 4 _ ⏟ 6 _ ⏟ 6  here total  6 × 6 = 36  possible cases So, total possible cases will be  5 + 24 + 180 + 216 + 108 + 36 = 569  numbers