The number of 6 digited numbers that can be formed having the property that every succeeding digits is greater than the preceding digit is equal to

$x_1<x_2<x_3<x_4<x_5<x_6$  when the number is $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$

Clearly no digit can be zero

Select 6 digits from $1,2,\mathrm{3........7},8,9$  in ${}^9{C}_6$  ways and place them in one order.

Total number of such numbers ${=}^9{C}_6\times 1{=}^9{C}_3$