While writing all the numbers from 700 to 1000, how many numbers occur in which the first digits greater than the second digit, and the second digit is greater than the third digit ?

When the second digit is 1, third digit can be 0, i.e., there is one such number.
When the second digit is 2, third digit can be 0 or 1, i.e., there are 2 such numbers.
When the second digit is 3, third digit can be 0, 1 or 2, i.e., there are 3 such numbers, and so on.
When the first digit is 7, second digit can be 1, 2, 3, 4, 5 or 6
So, there are (1 + 2 + 3 + 4 + 5 + 6) = 21 such numbers between 700 and 799
When the first digit is 8, second digit can be 1, 2, 3, 4, 5, 6 or 7
So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7) = 28 such numbers between 800 and 899
When the digit is 9, second digit can be 1, 2, 3, 4, 5, 6, 7 or 8
So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 such numbers between 900 and 999
Hence, the required number
= (21 + 28 + 36)
= 85