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?

To find the number of integers from 700 to 1000 where the hundred's digit (H) is greater than the ten's digit (T), and the ten's digit is greater than the unit's digit (U), we analyze the possible values for H. The condition is H > T > U. 1. For numbers from 700 to 799 (H=7): The condition becomes 7 > T > U. T must be chosen from {1, 2, 3, 4, 5, 6}. U must be chosen from {0, 1, ..., T-1}. This is equivalent to choosing 2 distinct digits from the set {0, 1, 2, 3, 4, 5, 6} and arranging them in decreasing order (the larger becomes T, the smaller becomes U). The number of ways to do this is 7C2 = (7 * 6) / 2 = 21. 2. For numbers from 800 to 899 (H=8): The condition becomes 8 > T > U. T must be chosen from {1, 2, ..., 7}. U must be chosen from {0, 1, ..., T-1}. This is equivalent to choosing 2 distinct digits from the set {0, 1, 2, 3, 4, 5, 6, 7}. The number of ways to do this is 8C2 = (8 * 7) / 2 = 28. 3. For numbers from 900 to 999 (H=9): The condition becomes 9 > T > U. T must be chosen from {1, 2, ..., 8}. U must be chosen from {0, 1, ..., T-1}. This is equivalent to choosing 2 distinct digits from the set {0, 1, 2, 3, 4, 5, 6, 7, 8}. The number of ways to do this is 9C2 = (9 * 8) / 2 = 36. 4. For the number 1000: The digits are H=1, T=0, U=0. The condition 1 > 0 > 0 is false because 0 is not greater than 0. So, 1000 does not satisfy the condition. Total numbers satisfying the condition = 21 (for H=7) + 28 (for H=8) + 36 (for H=9) = 85. Therefore, the correct answer is C) 85.