# Thread: code of credit cards

1. ## code of credit cards

how many codes of credits cards exist with only 2 algarisms diferents? algarisms: from 0 to 9

2. what do u mean by 2 algarisms diferents?
please give me at least an example, so I will try to solve it for u.
Also, credit card with sixteen zeros is it countable?

3. Originally Posted by hametceq
what do u mean by 2 algarisms diferents?
please give me at least an example, so I will try to solve it for u.
Also, credit card with sixteen zeros is it countable?
for example: 1122 or 1112, these numbers have 2 algarisms which are diferent.

No, only codes with 4 digits

4. Originally Posted by JoanF
how many codes of credits cards exist with only 2 algarisms diferents? algarisms: from 0 to 9
Originally Posted by JoanF
for example: 1122 or 1112, these numbers have 2 algarisms which are diferent.
No, only codes with 4 digits
It's not clear what you are asking.
In your example 1122 & 1112, are you looking for the manner in which the check digit is determined?
Each credit card company has it's own algorithm for it's credit card number.

I'm guessing that your word "algarism" means the same as "algorithm"

Can you supply some additional data?

5. Apparently, "algarism" is an arcane word for "digit", or "numeral". See, e.g., this Yahoo question or this article (written in English by Brazilians) that contains the following phrase: "The deviation on the fifth decimal algarism can be attributed to the accuracy of the numerical calculation." In fact, "algarismo" means "digit" in Portuguese.

6. Originally Posted by aidan
It's not clear what you are asking.
In your example 1122 & 1112, are you looking for the manner in which the check digit is determined?
Each credit card company has it's own algorithm for it's credit card number.

I'm guessing that your word "algarism" means the same as "algorithm"

Can you supply some additional data?

I only have this info

that's ok, thank's anyway

7. Well you need to choose two digits, so that can be done in 9*10 ways. Now you have 3 different cases. First digit goes in the code once, second twice; both of them go in the code twice; first goes three times.

You need to permute the digits for all of the three cases so in the end you have

9*10*(4!/3! + 4!/(2!*2!) + 4!/3!)

8. Originally Posted by Petkovsky
Well you need to choose two digits, so that can be done in 9*10 ways. Now you have 3 different cases. First digit goes in the code once, second twice; both of them go in the code twice; first goes three times.

You need to permute the digits for all of the three cases so in the end you have

9*10*(4!/3! + 4!/(2!*2!) + 4!/3!)
aren't you repeating when you do 4!/3! again?

I think it is only:
9*10*(4!/3! + 4!/(2!*2!))