How many six-digit integers can be formed having the property that every succeeding digit is greater than the preceding digit??
According to me, since the order of the digits is fixed....all we have to do is choose six distinct digits from the 10 digits we have.....and hence the answer to the above question should be equal to (10 C 6)......but my textbook mentions the answer as (9 C 3).....is my method incorrect? If yes then please suggest a way to solve.....Thanks in advance!!!
In the case that you can start with $0$, then ${{10}\choose{6}}$ is correct, as there are ${{10}\choose{6}}$ distinct digit selections (if the digits aren't distinct, then there is no correct sorting), and each selection has only one possible sorting.
If leading $0$s are not allowed, then the answer is simply ${{9}\choose{6}}$.