0

Limitations are: no repetitions, all lowercase(26) and must be in ascending order.

Now the main problem here is ascending order, so if you know any good resources that tackle this problem I would be very grateful. Also sorry if the questions has been asked, but I didn't find it :(

gx15
  • 1
  • Stars and bars. Google it. Basically the are 18 options for the first letter. If the first letter is k, there are 19-k options for the second, for rge third there are 20 - k - j and so on. If you do sumations of sumations you'll get somewhere. – fleablood Mar 04 '17 at 18:55
  • Oops. The answers are better than my cooment. Maybe I'll work out if my comment was wrong, or correct but harder than nescessary. – fleablood Mar 04 '17 at 18:57

2 Answers2

1

To choose such a password you have to choose $9$ letters from the set of $26$. You don't have to choose their order, because it is already chosen. The answer is $$\binom{26}9$$

ajotatxe
  • 65,084
0

Since there are no repetitions, we are simply choosing 9 distinct letters of the 26. Now, once we pick those 9, there is only one way to arrange them in ascending order, so the answer is just ${26}\choose{9}$.

Ziryerx
  • 554