If all the six digit numbers x _ 1 x _ 2 x _ 3 x _ 4 x _ 5 x _ 6...

Taking case $\left(1\right)$ fixing $1 \text{\text{ and } }$ 2 at first two places,

So, other number can be selected in $C4 7 = 35$ ways \left\{\text{as number are$from$1-9}\right\}$</p><p>Case \left(2\right)$ fixing $1 \text{\text{ and } }$ 3 at first two place,

So, other number can be selected in$C4 6 = 15$ ways

Similarly,

So, other number can be selected in$C4 5 = 5$ ways

Now fixing $1 \text{\text{ and } }$ 5 we get,

Other number can be selected in$C4 4 = 1$

Now fixing $2 \text{\text{ and } }$ 3 in first two place,

So, other number can be selected in $C4 6 = 15$

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

Hence, sum will be $2 + 4 + 5 + 6 + 7 + 8 = 32$