A : 0 exactly twice B : 1 exactly twice C:0 and 1 exactly twice ⇒n(A)‌=‌6C2⋅25[0‌ can't be leftmost digit ‌] ‌=480 B:(1 at leftmost place )∪(1 is not at leftmost place ) =‌6C1⋅25+‌6C2⋅24=432 C : 0011222, no. of valid 7 digit number ‌⇒‌
7!
3!2!2!
−‌
6!
3!2!
=150 ‌⇒‌ Total numbers ‌=480+432−150 ‌⇒762