Answer:
A
Explanation:
Correct option is A)
Let a 3-digit number ABC
Case (1)
If A=7, then B and C can be any digit but not 7
so, total 9 digits (0,1,2,3,4,5,6,8,9) can replace B and C
Then total such combinations =1(9)(9)=81
Case (2)
If B=7 then A can be any digit but not 0 and 7, because if A=0 it will be a 2-digit number.
And C can be any digit but not 7
Then total combinations =(8)(1)(9)=72
Case (3)
If C=7 then A can be any digit but not 0 and 7, because if A=0 it will be a 2-digit number.
And B can be any digit but not 7
Then total combinations =(8)(9)(1)=72
So, total numbers between 99 and 1000 which have exactly one of their digits as 7 are =81+72+72=225