Possible Combinations | ATM Pin

[skullers_ab]

Homework Statement

To gain access to his account, a customer using an automatic teller machine (ATM) must enter a four digit code. If repetition of the same four digit is not allowed (for example, 1111), how many possible combinations are there?

The Attempt at a Solution

I figured two possible solutions. I do not know which one is correct and why the other one is not.

Possibility 1:
Considering 3 digits which can have 10 combinations each (0-9) and the last digit which can only have 9 combinations, to prevent the repetitions of all 4 digits; we have

10 X 10 X 10 X 9 = 9000

Possibility 2:
Considering that the number of combinations that are not allowed are 10 (0000, 1111, 2222, ..., 9999), we just take the total no. of possible combinations and subtract 10 from it. Thus,

(10^4) - 10 = 9990

I am biased towards the 2nd possibility but do not understand why the first one is wrong.
Explanation/help would be very much appreciated.

 

[HallsofIvy]

Your "last digit which can only have 9 combinations, to prevent the repetitions of all 4 digits" would allow things like 1119 but NOT 1911 which is also a valid combination. There is nothing special about the last digit.

 

[skullers_ab]

Pardon me if I misunderstood your point.

I didn't mean the last digit in the sense as the 4th digit. I meant that at least one of the digits has to be different and thus in any PIN, at least one of the digits will not be repeated, and thus will only have a selection of 9 numbers to choose from and not 10, unlike the other digits which will. Hence, 1119 and 1911 are both accounted for by 10 X 10 X 10 X 9, (which is the same as 10 X 9 X 10 X 10)

I'm sorry if I am not advocating the fact. My knowledge in combinatorics is poor and my expression bad. Intuition is a major factor in my point above.

If you are sure that what I'm saying is wrong, please let me know once more and I'll accept it and try to understand the topic better by reading more later on. Thanks for the reply.

 

[Kaimyn]

Pardon me if I misunderstood your point.

I didn't mean the last digit in the sense as the 4th digit. I meant that at least one of the digits has to be different and thus in any PIN, at least one of the digits will not be repeated, and thus will only have a selection of 9 numbers to choose from and not 10, unlike the other digits which will. Hence, 1119 and 1911 are both accounted for by 10 X 10 X 10 X 9, (which is the same as 10 X 9 X 10 X 10)

I'm sorry if I am not advocating the fact. My knowledge in combinatorics is poor and my expression bad. Intuition is a major factor in my point above.

If you are sure that what I'm saying is wrong, please let me know once more and I'll accept it and try to understand the topic better by reading more later on. Thanks for the reply.

What about things like $$123x$$ where x is a digit? "x" could be any of the ten* possibilities, yet you are restricting it to nine. You are eliminating options that work.

*1230,1231,1232,1233,1234,1235,1236,1237,1238,1239


Your second option is correct. You get it by having the total number of possibilities, including the incorrect ones and getting rid of said options.
Obviously, the digits must be the same, so every subsequent digit will be the same as the first. So we get

$$10\times10\times10\times10 - 10\times1\times1\times1$$
Which reduces to your second option.

 

[skullers_ab]

Hmmm...that does make sense.

Earlier I was unable to understand which valid numbers I was 'throwing away' in my first option and why they were being 'thrown away'.

Thank you for very much for the explanation. It's quite clear to me now.

Cheers

 

[Kaimyn]

No problem