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While writing all the numbers from 700 to 1000, how many numbers occur in which the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place?
- a)61
- b)64
- c)85
- d)91
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
While writing all the numbers from 700 to 1000, how many numbers occur...
's place is greater than or equal to the digit at ten's place?
There are two possible cases for the digits at hundred's and ten's place: either the hundred's digit is greater than or equal to the ten's digit, or it is less than the ten's digit.
Case 1: Hundred's digit is greater than or equal to ten's digit
In this case, we can choose any digit from 7 to 9 for the hundred's place, and any digit from the hundred's digit to 9 for the ten's place. For example, if we choose 8 as the hundred's digit, we can choose any digit from 8 to 9 for the ten's digit. Therefore, there are 3 choices for the hundred's digit and for each choice, there are (10 - hundred's digit + 1) choices for the ten's digit. So, the total number of numbers in this case is:
3 × [(10 - 7 + 1) + (10 - 8 + 1) + (10 - 9 + 1)]
= 3 × (4 + 3 + 2)
= 27
Case 2: Hundred's digit is less than ten's digit
In this case, we can choose any digit from 7 to 8 for the hundred's place, and any digit from 0 to 9 for the ten's place, except for the digit chosen for the hundred's place. For example, if we choose 7 as the hundred's digit, we can choose any digit from 0 to 9 for the ten's digit, except for 7. Therefore, there are 2 choices for the hundred's digit, and for each choice, there are 9 choices for the ten's digit. So, the total number of numbers in this case is:
2 × 9
= 18
Therefore, the total number of numbers in which the hundred's digit is greater than or equal to the ten's digit is:
27 + 18
= 45
Therefore, there are 45 numbers in which the digit at hundred's place is greater than or equal to the digit at ten's place.
There are two possible cases for the digits at hundred's and ten's place: either the hundred's digit is greater than or equal to the ten's digit, or it is less than the ten's digit.
Case 1: Hundred's digit is greater than or equal to ten's digit
In this case, we can choose any digit from 7 to 9 for the hundred's place, and any digit from the hundred's digit to 9 for the ten's place. For example, if we choose 8 as the hundred's digit, we can choose any digit from 8 to 9 for the ten's digit. Therefore, there are 3 choices for the hundred's digit and for each choice, there are (10 - hundred's digit + 1) choices for the ten's digit. So, the total number of numbers in this case is:
3 × [(10 - 7 + 1) + (10 - 8 + 1) + (10 - 9 + 1)]
= 3 × (4 + 3 + 2)
= 27
Case 2: Hundred's digit is less than ten's digit
In this case, we can choose any digit from 7 to 8 for the hundred's place, and any digit from 0 to 9 for the ten's place, except for the digit chosen for the hundred's place. For example, if we choose 7 as the hundred's digit, we can choose any digit from 0 to 9 for the ten's digit, except for 7. Therefore, there are 2 choices for the hundred's digit, and for each choice, there are 9 choices for the ten's digit. So, the total number of numbers in this case is:
2 × 9
= 18
Therefore, the total number of numbers in which the hundred's digit is greater than or equal to the ten's digit is:
27 + 18
= 45
Therefore, there are 45 numbers in which the digit at hundred's place is greater than or equal to the digit at ten's place.
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Community Answer
While writing all the numbers from 700 to 1000, how many numbers occur...
The correct option is C.
For numbers from 700 to 1000, the digit at hundreds place can be 7, 8 or 9. So, three possibilities are there.
Now, for the digit at ten's place to be greater than the digit at the unit's place, total numbers are equal to the value of tens digit. For example, for 2 at ten's digit, numbers can be 20, 21 (total 2). Similarly, for 6 at ten's place, numbers can be 60, 61, 62, 63, 64, 65 (total 6).
When the first digit is 7, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 = 21
When the first digit is 8, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
When the first digit is 9, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36
Therefore, required total numbers = 21 + 28 + 36 = 85.
For numbers from 700 to 1000, the digit at hundreds place can be 7, 8 or 9. So, three possibilities are there.
Now, for the digit at ten's place to be greater than the digit at the unit's place, total numbers are equal to the value of tens digit. For example, for 2 at ten's digit, numbers can be 20, 21 (total 2). Similarly, for 6 at ten's place, numbers can be 60, 61, 62, 63, 64, 65 (total 6).
When the first digit is 7, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 = 21
When the first digit is 8, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
When the first digit is 9, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36
Therefore, required total numbers = 21 + 28 + 36 = 85.
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