Total number of six-digit numbers that can be formed having the property that every succeeding digit is greater than the preceding digit is equal to 

${\mathrm{x}}_1<{\mathrm{x}}_2<{\mathrm{x}}_3<{\mathrm{x}}_4<{\mathrm{x}}_5<{\mathrm{x}}_6\text{, }$when the number is ${\text{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3{\mathrm{x}}_4{\mathrm{x}}_5{\mathrm{x}}_6$
Clearly no digit can be zero. Also, all the digits are distinct. So, let us first select six digits from the list of digits 1,2,3,4, 5, 6, 7 , 8,9 which can be done in ${ }^9{C}_6$ ways.
After selecting these digits they can be put only in one order. Thus, total number of such numbers is ${ }^9{C}_6\times 1{=}^9{C}_3$.