How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number.
There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $$9C_2$$ ways.
3 digit numbers can be formed in $$9C_3$$ ways.
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..9 digit number can be formed in 9C9 ways. 

We know that $$nC_0+nC_1+nC_2+.........+nC_n =2^n$$
=> $$9C_0 + 9C_1+9C_2+....9C_9 = 2^9$$
$$9C_0 + 9C_1 + ...9C_9 = 512$$

We have to subtract $$9C_0$$ and $$9C_1$$ from both the sides of the equations since we cannot form single digit numbers. 
=> $$9C_2 + 9C_3+...+9C_9=512-1-9$$
$$9C_2+9C_3+...+9C_9=502$$

Therefore, $$502$$ is the right answer.