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I need to know the total possible unique variations there can be on an identifier that is made up of 20 alphanumeric characters, where the characters are A to Z (all upper case), and the digits 0 to 9.

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    13.36 nonillion2012-09-27

2 Answers 2

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If by this question you mean the number of strings of length $20$ from the alphabet $\{ A, B, \cdots, Z, 0, 1, \cdots, 9 \}$, then: there are $36$ letters in this alphabet, and you can choose any one of them for each of the $20$ characters in the string, so that leaves $36^{20}$ possible strings.

That's a lot of strings. $13\,367\,494\,538\,843\,734\,067\,838\,845\,976\,576$ to be more precise.

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    @BrianMScott: Or even $13367494538843734067838845976576$ ;) Wolfram Alpha has its uses...2012-09-27
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Actually, it is higher: $13,749,422,954,239,300,000,000,000,000,000$. As you have UP TO $20$ characters, you need to take into account all the possibilities prior to the max, i.e. $36 \\ 1,296 \\ 46,656 \\ 1,679,616 \\ 60,466,176 \\ 2,176,782,336 \\ 78,364,164,096 \\ 2,821,109,907,456 \\ 101,559,956,668,416 \\ 3,656,158,440,062,980 \\ 131,621,703,842,267,000 \\ 4,738,381,338,321,620,000 \\ 170,581,728,179,578,000,000 \\ 6,140,942,214,464,820,000,000 \\ 221,073,919,720,733,000,000,000 \\ 7,958,661,109,946,400,000,000,000 \\ 286,511,799,958,070,000,000,000,000 \\ 10,314,424,798,490,500,000,000,000,000 \\ 371,319,292,745,659,000,000,000,000,000 \\ 13,367,494,538,843,700,000,000,000,000,000$

Total sum: $13,749,422,954,239,300,000,000,000,000,000$.

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    @ArthurFischer: Haha, my mistake. I had misinterpreted it as a mistaken version of "*made of up to 20* alphanumeric characters".2015-08-28