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 ✓ UPSC's answer
(d) 91

Why the answer is (c)

• The numbers range from 700 to 1000, so the hundreds digit (H) can be 7, 8, or 9.

• The condition is H > T > U, where T is the tens digit and U is the units digit.

• For H=7, T can be 6, 5, 4, 3, 2, 1, 0. For each T, U can be any digit from 0 to T-1. The count is sum of (T) for T=0 to 6, which is 6+5+4+3+2+1+0 = 21.

• For H=8, T can be 7, 6, 5, 4, 3, 2, 1, 0. The count is sum of (T) for T=0 to 7, which is 7+6+5+4+3+2+1+0 = 28.

• For H=9, T can be 8, 7, 6, 5, 4, 3, 2, 1, 0. The count is sum of (T) for T=0 to 8, which is 8+7+6+5+4+3+2+1+0 = 35.

• Total count = 21 + 28 + 35 = 84. Wait, let me re-evaluate the summation logic. The number of choices for U given T is T (digits 0 to T-1). So for H=7, T ranges 0-6. Sum = 0+1+2+3+4+5+6 = 21. For H=8, T ranges 0-7. Sum = 0+1+...+7 = 28. For H=9, T ranges 0-8. Sum = 0+1+...+8 = 36. Total = 21 + 28 + 36 = 85.

Why the other options are wrong

(a) 61: 61 is incorrect because it likely results from an undercount of the valid combinations for the tens and units digits.
(b) 64: 64 is incorrect because it does not match the calculated sum of valid permutations for hundreds digits 7, 8, and 9.
(d) 91: 91 is incorrect because it overcounts the possibilities, possibly by including cases where the hundreds digit is not strictly greater than the tens digit.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2018, held on 3 June 2018. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.