To find the total no. of six digit numbers that can be formed having property that every succeeding digit is greater than preceding digit. [closed]

Solution 1:

Hint: Once you've picked six different digits, there is only one way to arrange them to satisfy the conditions given.

Solution 2:

Or to look at it another way. There is only one 9 digit number with this condition 123456789.

What about 8 digits?: there are nine answers 12345678, 12345679, 12345689, 12345789, 12346789, 12356789, 12456789, 13456789, 23456789.

So all I have done is strike out each digit in turn. $9\times 1$ digit = 9 possibilities. For 7 digits, you need to strike out 2 digits - how many ways of doing this are there? does $\binom{9}{2}$ ring a bell? And so for 6 digit number you need to lose 3 digits.........