Question

How many numbers are there between $ 100 $ and $ 1000 $ such that at least one of their digits is $ 7 $ ?

Answer 1

Hint:
For this problem, we will use the concept of permutation and combinations. By using this concept, we will calculate the number of possible ways to make a $ 3 $ digit number which lies between $ 100 $ and $ 1000 $ such that at least one of their digits is $ 7 $. For this, we will consider three cases. The first one is “the number having only one $ 7 $ ”, the second one is “the number having two $ 7 $ ’s”, and the third case is “the number having all three digits as $ 7 $ ”. We will add all of them to get the result.

Complete step by step answer:
Assuming a three digited number, which lies between $ 100 $ and $ 1000 $.
Case-1: Only one digit is $ 7 $ .
(i) When the one’s place is $ 7 $ .
The one’s place can be arranged in one way, tens place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways to get a number greater than $ 100 $ we neglected $ 0 $ in the tens place, hundreds place can be filled by $ (2,3,4,5,6,7,8,9) $ in $ 8 $ ways to get a number greater than $ 100 $ , we neglected $ 0,1 $ in the hundreds place.
Hence the total number of ways is $ 1\times 9\times 8=72 $ .
(ii) When the tens place is $ 7 $ .
The one’s place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways, tens place is only filled by $ 7 $ in one way, hundreds place can be filled by $ (2,3,4,5,6,7,8,9) $ in $ 8 $ ways to get a number greater than $ 100 $ , we neglected $ 0,1 $ in the hundreds place.
Hence the total number of ways is $ 1\times 9\times 8=72 $ .
(iii) When the hundreds place is $ 7 $ .
The ones place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways, tens place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways to get a number greater than $ 100 $ we neglected $ 0 $ in tens place, hundreds place can be filled by $ 7 $ in one way.
Hence the total number of ways is $ 1\times 9\times 9=81 $ .
Case-2: When two digits are $ 7 $ .
(i) when ones and tens place are filled with $ 7 $ .
Ones and tens place are filled with $ 7 $ in only one way, hundreds place can be filled by $ (2,3,4,5,6,7,8,9) $ in $ 8 $ ways to get a number greater than $ 100 $ , we neglected $ 0,1 $ in the hundreds place.
Hence the total number of ways is $ 1\times 8=8 $ .
(ii) When hundreds and tens place is $ 7 $ .
The one’s place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways, tens and hundreds place can be filled by $ 7 $ in only one way.
Hence the total number of ways is $ 1\times 9=9 $ .
(iii) When ones and hundred place is $ 7 $ .
The ones and hundreds place can be filled by $ 7 $ in one way, tens place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways.
Hence the total number of ways is $ 1\times 9=9 $ .
Case-3: When all three digits are $ 7 $ .
We have only one possible way to fill the $ 7 $ in all three digits.
Hence the total number of ways is $ 1 $ .
Now all the possible ways to form a digit which is having at least one digit as $ 7 $ between $ 100 $ and $ 1000 $ is $ 72+72+81+8+9+9+1=252 $ .

Note:
 We can simply solve this problem by calculating the number of possible ways to form a number which do not have $ 7 $ as a digit between $ 100 $ and $ 1000 $. We can fill ones, tens place in $ 9 $ ways by $ (1,2,3,4,5,6,7,8,9) $ and hundred places can be filled in $ 8 $ ways by $ (2,3,4,5,6,7,8,9) $ , hence the possible ways are $ 8\times 9\times 9=648 $ . Now the required value can be obtained by subtracting the calculated value from a number of numbers between $ 100 $ and $ 1000 $. We have $ 1000-100=900 $ numbers between $ 100 $ and $ 1000 $ .
 $ \begin{align}
  & \therefore x=900-648 \\
 & \Rightarrow x=252 \\
\end{align} $
From both the methods we got the same result.