If all the six digit numbers x_1x_2x_3x_4x_5x_6 with 0

Question

If all the six digit numbers $x_1x_2x_3x_4x_5x_6$ with $0<x_1<x_2<x_3<x_4<x_5<x_6$ are arranged in increasing order, then the sum of the digits in the 72nd number is \_\_\_\_\_\_\_.

Answer: 32

Accepted Answer

Numbers starting with 1: $\binom{8}{5}=56$ such numbers (choosing remaining 5 digits from 2--9). Numbers starting with 2: $\binom{7}{5}=21$; these bring the cumulative count to $56+21=77$, so the $72^{\text{nd}}$ number starts with 2 and is the $(72-56)=16^{\text{th}}$ number starting with 2.

Continuing the digit-by-digit ranking within "2xxxxx" leads to the number $245678$ as the $72^{\text{nd}}$ number.

$$\text{Sum of digits} = 2+4+5+6+7+8 = 32$$

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