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

Last updated date: 09th Sep 2024
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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.

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.