number of combination of 8 digits number

• Apr 6th 2008, 10:04 AM
lekge
number of combination of 8 digits number
Mr. Azzamer wants to invent a computer password of 8 symbols long with the following properties:
i) each symbol is one of the digits {1, 2, 3, 4, 5, 6, 7, 8, 9}
ii) the first digit of the password is at least 1, the second digit is at least 2, the third digit is at least 3 and so on.
iii) no digit can be used more than once in the same password.
• Apr 6th 2008, 02:12 PM
roy_zhang
Quote:

Originally Posted by lekge
Mr. Azzamer wants to invent a computer password of 8 symbols long with the following properties:
i) each symbol is one of the digits {1, 2, 3, 4, 5, 6, 7, 8, 9}
ii) the first digit of the password is at least 1, the second digit is at least 2, the third digit is at least 3 and so on.
iii) no digit can be used more than once in the same password.

If we think about the possibilities of the passwords from right to left (i.e. let's think about how many different digits we can choose as the eighth digit in the password first), by the requirements, we have 2 digits can be chosen as the eighth digit (i.e 8 or 9). Now let's move one space to the left, how many different digits we can choose as the seventh digit? The candidates are 7,8 or 9, but one of these three digits already been chosen as the eighth digit, so we have just 2 digits left to be chosen as the seventh digits. If we proceed this way all the back to the first digit, we found out the the total number of possible passwords is just $2^8=256$.

Roy
• Apr 6th 2008, 10:23 PM
lekge
Quote:

Originally Posted by roy_zhang
If we think about the possibilities of the passwords from right to left (i.e. let's think about how many different digits we can choose as the eighth digit in the password first), by the requirements, we have 2 digits can be chosen as the eighth digit (i.e 8 or 9). Now let's move one space to the left, how many different digits we can choose as the seventh digit? The candidates are 7,8 or 9, but one of these three digits already been chosen as the eighth digit, so we have just 2 digits left to be chosen as the seventh digits. If we proceed this way all the back to the first digit, we found out the the total number of possible passwords is just $2^8=256$.

Roy

sorry to tell u that your solution not that true, for example, if u taking 8th digit as 8, as u mention above, 7th digit only can taken from {7,8,9},but 8,9 is not possible for 7th digit, hence only 7 is allow for 7th digit. Contradiction that there got 2 choice.

i found the solution from 8 个数字排法。_百度知道 by the user "lris_lucky"

method 1:
"123456789"
since the sequence must in increasing order,
if we take out a number from the number above, or taking 8 number from the number above, we are done. Hence solution should be c(9,1)=c(9,8)=9

method 2:
"12345678" is lowest and "23456789" is biggest.

when the first digit is 1, the other 7 digits got c(8,7)=8 ways of combination, however if the first digit is 2, there only exist a combination "23456789".
therefore the total combination
= c(8,7)+1
=8+1
=9
• Apr 7th 2008, 01:06 AM
angel.white
Can someone else answer this question? I think that roy is correct, but I'm not confident enough in my understanding of probability to post my reasoning.
• Apr 7th 2008, 03:37 AM
Plato
Quote:

Originally Posted by lekge
since the sequence must in increasing order

That requirement does not appear in the original statement of the problem.

Quote:

Originally Posted by roy_zhang
we found out that the total number of possible passwords is just $2^8=256$.

Without the above requirement roy_zhang抯 answer is correct.