If all the six digit numbers x₁ x₂ x₃ x₄ x₅ x₆ with 0< x₁ < x₂ < x₃ < x₄ < x₅ < x₆ are arranged in the increasing order, then the sum of the digits in the 72ᵗʰ number is _____________.

Solution

Maths sir · 3 steps

Each number is just a choice of 6 digits from 1 to 9, so count the blocks by first digit and walk to the 72nd.

1. Choose, do not arrange. The digits must be strictly increasing and nonzero, so every such number is a choice of 6 digits from {1, 2, …, 9}, written in increasing order. Each choice gives exactly one number.

   Numbers starting with	Pick the rest from	Count	Running total
   1	5 digits from {2, …, 9}	C(8, 5) = 56	56
   2, 3	4 digits from {4, …, 9}	C(6, 4) = 15	71
   2, 4	4 digits from {5, …, 9}	C(5, 4) = 5	76

2. Locate the 72nd. The numbers starting with 1, and then those starting with 2, 3, fill positions 1 to 71. So the 72nd number is the smallest number that starts with 2, 4, which is 245678.
3. So. The digit sum is 2 + 4 + 5 + 6 + 7 + 8 = 32.

The common slip: 35. Taking the last number of the 2, 3 block (236789, which is the 71st) gives a digit sum of 35, not 32.

The idea to keep. When digits must be strictly increasing, order is fixed, so counting is pure C(n, r). Fix digits from the left, add block sizes, and stop when the running total crosses the target rank.

Answer from the NTA final answer key. Solution by PrepWiser.