TITLE: To find the total no. of six digit numbers that can be formed having property that every succeeding digit is greater than preceding digit.
QUESTION [5 upvotes]: I have a question and got strucked on this..
To find the total no. of six digit numbers that can be formed having property that every succeeding digit is greater than preceding digit.
Please guide me to solve this. Thanks in advance.

REPLY [7 votes]: 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.........