Question

How many numbers between 100 and 1000 are such that exactly one of the digits is 6?

Answer 1

Hint:
Here, we need to find how many numbers between 100 and 1000 are such that exactly one of the digits is 6. We will find the count of numbers that have 6 at the hundred’s place, the count of numbers that have 6 at the ten’s place, and the count of numbers that have 6 at the unit’s place. Finally, we will add the three cases to find the count of numbers between 100 and 1000 that have exactly one of the digits as 6.

Complete step by step solution:
We need to count the numbers between 100 and 1000 which have exactly one of the digits as 6.
All the numbers between 100 and 1000 are three digit numbers.
Therefore, exactly one of the three digits must be 6.
We will use three cases to find the required answer.
Case 1: The number 6 is at the hundred’s place.
The three digit number will have only one 6. Therefore, the remaining digits are 0, 1, 2, 3, 4, 5, 7, 8, 9.
In the ten’s place, any of the remaining 9 numbers (except 6) can be placed.
In the unit’s place, any of the remaining 9 numbers (except 6) can be placed.
Therefore, we get
\[ \Rightarrow \]Number of three digit numbers with exactly one 6 (at the hundred’s place) \[ = 9 \times 9\]
Multiplying the terms, we get
\[ \Rightarrow \]Number of three digit numbers with exactly one 6 (at the hundred’s place) \[ = 81\] numbers
Case 2: The number 6 is at the ten’s place.
The three digit number will have only one 6. Therefore, the remaining digits are 0, 1, 2, 3, 4, 5, 7, 8, 9.
We know that a three digit number cannot have the digit 0 in the hundred’s place.
Thus, in the hundreds place, any of the remaining 8 numbers (except 0 or 6) can be placed.
In the unit’s place, any of the remaining 9 numbers (except 6) can be placed.
Therefore, we get
\[ \Rightarrow \]Number of three digit numbers with exactly one 6 (at the ten’s place) \[ = 8 \times 9\]
Multiplying the terms, we get
\[ \Rightarrow \]Number of three digit numbers with exactly one 6 (at the ten’s place) \[ = 72\] numbers
Case 3: The number 6 is at the unit’s place.
The three digit number will have only one 6. Therefore, the remaining digits are 0, 1, 2, 3, 4, 5, 7, 8, 9.
We know that a three digit number cannot have the digit 0 in the hundred’s place.
Thus, in the hundreds place, any of the remaining 8 numbers (except 0 or 6) can be placed.
In the ten’s place, any of the remaining 9 numbers (except 6) can be placed.
Therefore, we get
\[ \Rightarrow \]Number of three digit numbers with exactly one 6 (at the unit’s place) \[ = 8 \times 9\]
Multiplying the terms, we get
\[ \Rightarrow \]Number of three digit numbers with exactly one 6 (at the unit’s place) \[ = 72\] numbers
Finally, we will calculate the total numbers between 100 and 100 which have exactly one digit as 6.
The count of numbers between 100 and 100 which have exactly one digit as 6 is the sum of the numbers having 6 at hundreds place, the numbers having 6 at ten’s place, and the numbers having 6 at unit’s place.
Therefore, we get the count of numbers as \[81 + 72 + 72 = 225\] numbers.

Thus, there are 225 numbers between 100 and 1000 that have exactly one of the digits as 6.

Note:
A common mistake is to consider 0 at hundreds places also in case 2 and case 3. This is incorrect because the number 068 is the same as 68, which is not a three digit number. It does not lie between 100 and 1000.