# How many permutations?

I'm trying to calculate the number of possible non-repeated permutations of these serial key styles.

If someone could either let me know how many for each, or how I might calculate it, I'd be very grateful.

The character set is 24 uppercase letters, and 8 numbers. (removing I, O, 0, 1)

1) 7FB-E48-W60

2) J5VR-CN4E-GWG8

3) Q5YJ-NMZN-RF48-GSW4

4) H6EFA-N6H7O-08WW8-0S4SC-4K4S8


Thanks very much.

Mike.

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You say "removing I, O, 0, 1$, but I see 0,1, and O in your examples. Do you mean these are examples without the restrictions? – Thomas Andrews Mar 13 '12 at 12:39 Hi, Thomas. Sorry, these examples don't have those excluded, they're to demonstrate the grouping and length. The finished product would have those excluded. – i-CONICA Mar 13 '12 at 13:02 Hi, dtldarek. I struggle with simple multiplication and division when it involves odd numbers. I don't know what any of that stuff means. I was hoping for pointers on what to type into a calculator, or just the answer for each. – i-CONICA Mar 13 '12 at 13:07 To be specific, what do you mean by "non-repeated" and what do you mean by "permutations of styles"? – dtldarek Mar 13 '12 at 13:08 Hi, Sorry that part was unclear. I'm choosing to adopt one of these, numbered 1, to 4 styles of codes to use as voucher codes to be printed on coupon cards to be sold in stores. I want to know how many of these "keys" or "serials" I could generate uniquely with the character set of 24 upper case letters and 8 numbers, by "non-repeated" I mean unique entire keys, but each key can contain the same character more than once. I'd prefer to use style 1, but if that only gives me 2 million possible serial keys, then I'd have to go for 2, etc... – i-CONICA Mar 13 '12 at 13:15 ## 4 Answers 1. The dashes don't make any difference. 2. Your alphabet has 24 + 8 = 32 characters. 3. There are 32^n = 32 * 32 * 32 * ... * 32 * 32 (n times) different strings of length n using this alphabet. 4. Using your schemes: 1. 32^9 = 35184372088832, 2. 32^12 = 1152921504606846976, 3. 32^16 = 1208925819614629174706176, 4. 32^25 = 42535295865117307932921825928971026432. Have fun ;-) - since without repetition: (32 because 8+24) 1- 32*31*30*29*28*27*26*25*24 =$\frac{32!}{(32-9)}!$2-$\frac{32!}{(32-12)}!$3-$\frac{32!}{(32-16)}!$4-$\frac{32!}{(32-25)}!$ But your example above shows repetition and I,O,0, & 1. I hope that is what you want. - Hi, Zeina. Sorry, I written the question a little confusingly. By "non-repeating" I mean how many of these keys could I generate uniquely. Each key can contain the same character more than once. I've left some notes on the question. Thank you for your time. – i-CONICA Mar 13 '12 at 13:17 If, as you say in an earlier comment, characters can occur multiple times in a given serial number, then it's quite straightforward: the total number of serial numbers available for a key of length$k$is$32^k$, where 32 is the size of your character set (24 letters and 8 digits). In practice it's usually more complicated, though, since the serial key is generated in such a way that some information can be extracted from the key (most notably whether or not the key is valid, but also things like an identifier for the product and its version number). - In that case the probabilities become: 1-$\32^{9} = 35184372088832

2- $\32^{12} 3-$\32^{16}

4- \$\32^{25}

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