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?

### Answer

As the digits appear in ascending order in the numbers, number of ways of forming a n-digit number using nine digits = ^{9}C_{n}

Number of possible two-digit numbers which can be formed

= ^{9}C_{2} + ^{9}C_{3} + ^{9}C_{4} + ^{9}C_{5} + ^{9}C_{6} + ^{9}C_{7} + ^{9}C_{8} + ^{9}C_{9}

= 2^{9} − (^{9}C_{1} + ^{9}C_{1})

= 512 − (1 + 9)

= 502

**The correct answer is 502.**