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?
Answers
Answer:
85
Step-by-step explanation:
Given 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?
Now for the digit at ten's place is greater than the digit at unit's place, total numbers is equal to value of ten's place. For 3 at ten's place numbers can be 30, 31.
In the same way for 5 at ten's place numbers can be 50, 51, 52, 53, 54, 55 (6 numbers)
From 700 to 800, the first number is 7, so we get
6 + 5 + 4 + 3 + 2 + 1 = 21 numbers
From 801 to 900, he first number is 8, so we get
7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 numbers
From 901 to 1000, he first number is 8, so we get
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 numbers
Adding all the numbers we get the total as
21 + 28 + 36 = 85 numbers