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
Show answer and explanation
Correct answer: C - 85
We need three-digit numbers with strictly decreasing digits (hundreds > tens > units). The number 1000 has digits 1,0,0,0 and does not qualify, so the hundreds digit can only be 7, 8 or 9. For a fixed hundreds digit H, the tens and units digits must be two distinct digits both less than H, arranged in decreasing order; choosing any 2 distinct digits from {0,1,...,H-1} fixes their order, giving C(H, 2) numbers. For H = 7: C(7,2) = 21. For H = 8: C(8,2) = 28. For H = 9: C(9,2) = 36. Total = 21 + 28 + 36 = 85. Hence (c) 85.