If all the six digit numbers x 1 x 2 x 3 x 4 x 5 x 6 with 0 < x 1 < x 2 < x 3 < x 4 < x…

Taking case $\left(1\right)$ fixing $1 \& 2$ at first two places,

So, other number can be selected in ${}^7\mathrm{C}_4=35$ ways $\left\{\text{as number are from 1-9}\right\}$

Case $\left(2\right)$ fixing $1 \& 3$ at first two place,

So, other number can be selected in${}^6\mathrm{C}_4=15$ ways

Similarly,

So, other number can be selected in${}^5\mathrm{C}_4=5$ ways

Now fixing $1 \& 5$ we get,

Other number can be selected in${}^4\mathrm{C}_4=1$

Now fixing $2 \& 3$ in first two place,

So, other number can be selected in ${}^6\mathrm{C}_4=15$

So, the ${72}^{nd}$ number will be $245678$

Hence, sum will be $2+4+5+6+7+8=32$