Answer
414.6k+ views
Hint:
we know about the units’ place, tens place, and hundreds place. If we take units place as 5, then we can take tens place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Complete step by step solution:
The number of numbers between 100 and 1000 such that exactly one of the digits is 5.
When the units place has 5, it means we can take tens place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Hence, there are $9 \times 8 = 72$ numbers.
When the tens place has 5, it means we can take unit place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Hence, there are $9 \times 8 = 72$ numbers.
When the hundreds place has 5, it means we can take units place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take tens place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled in 9 different ways.
Hence, there are $9 \times 9 = 81$ numbers.
Now, calculate the required numbers of numbers is:
$72 + 72 + 81 = 225$
Hence, the number of numbers between 100 and 1000 is 225. The option (B) is the correct option.
Note:
Here you should know each digit has 10 choices from 0 to 9 for every range of numbers. The unit place, tens place and hundred place must be taken with the required conditions. Make sure to remember that the problem says with repetition.
we know about the units’ place, tens place, and hundreds place. If we take units place as 5, then we can take tens place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Complete step by step solution:
The number of numbers between 100 and 1000 such that exactly one of the digits is 5.
When the units place has 5, it means we can take tens place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Hence, there are $9 \times 8 = 72$ numbers.
When the tens place has 5, it means we can take unit place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Hence, there are $9 \times 8 = 72$ numbers.
When the hundreds place has 5, it means we can take units place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take tens place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled in 9 different ways.
Hence, there are $9 \times 9 = 81$ numbers.
Now, calculate the required numbers of numbers is:
$72 + 72 + 81 = 225$
Hence, the number of numbers between 100 and 1000 is 225. The option (B) is the correct option.
Note:
Here you should know each digit has 10 choices from 0 to 9 for every range of numbers. The unit place, tens place and hundred place must be taken with the required conditions. Make sure to remember that the problem says with repetition.
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