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 the increasing order, then the sum of the digits in the $$72^{th}$$ number is ______.

We consider six-digit numbers $$x_1x_2x_3x_4x_5x_6$$ with $$0 < x_1 < x_2 < x_3 < x_4 < x_5 < x_6$$ arranged in increasing order, and we wish to find the sum of digits of the 72nd number.

Since $$x_1>0$$, the digits are chosen from $$\{1, 2, \ldots, 9\}$$ and all six must be strictly increasing, giving a total of $$\binom{9}{6} = 84$$ such numbers. Counting by the value of $$x_1$$ shows that there are $$\binom{8}{5}=56$$ numbers starting with $$x_1=1$$, $$\binom{7}{5}=21$$ starting with $$x_1=2$$, $$\binom{6}{5}=6$$ starting with $$x_1=3$$, and $$\binom{5}{5}=1$$ starting with $$x_1=4$$.

Therefore, the first 56 numbers have $$x_1=1$$, so the numbers in positions 57 through 77 have $$x_1=2$$. Hence the 72nd number is the $$72-56=16$$th number among those with $$x_1=2$$.

Within this group, if $$x_2=3$$ there are $$\binom{6}{4}=15$$ numbers obtained by choosing the remaining four from $$\{4,\ldots,9\}$$, and if $$x_2=4$$ there are $$\binom{5}{4}=5$$ numbers from $$\{5,\ldots,9\}$$. Since 16 exceeds 15, the 16th such number has $$x_2=4$$ and is the first in this subgroup, namely $$2,4,5,6,7,8$$.

Therefore, the 72nd number is $$245678$$, and the sum of its digits is $$2+4+5+6+7+8=32$$.