How do I go about finding how many different number/letter combinations are possible with the 10-digit codes on the iTunes Pepsi bottles?
March 1, 2004 11:07 PM
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How do I go about finding how many different number/letter combinations are possible with the 10-digit codes on the iTunes Pepsi bottles? [more]
Is it 10^36? The code is made up of ten letters or number which, in my experience, don't have to only appear once in the code. 0-9 and A-Z makes that 36 possible letters or numbers in each of the ten spaces.
If I haven't just answered my own question, should I instead be thinking about factorials? Or am I simply remembering something from Algebra II that's useless in this case?
There could be as many as 100 million winning codes, assuming Pepsi and Apple planned for that many winners. I just wanted to figure out in the grand scheme of things how many codes were theoretically possible.
posted by emelenjr to grab bag (11 comments total)
We know that for three digits the lowest number is 000 and the highest is 999, and we know for certain that means there's 1000 combinations. Which is... 10 ^ 3.
So, the answer is (options) ^ (numbers), or 36 ^ 10.
This can also be shown for binary numbers. It is well known an 8 bit number has 256 combinations, which is 2 ^ 8.
(And I wonder how I passed math, too)
Factorials are for figuring out things where you remove previous possibilities, like playing lotteries, IIRC.
posted by shepd at 11:22 PM on March 1, 2004