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Fabman
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farful
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Fabman
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Fabman
But there can only be 1 out come on each die per roll

I'm not following.


Nevertheless, with combinatorial problems, the best thing to do is to first simplify the problem, figure out a formula, then scale up.

Let's say we have a three 3-sided dice (yes, such a thing doesn't exist, whatever!).
Die 1 has words A, B, C, die 2 has words D, E, F, and die 3 has words G, H, I.

Convince yourself (by writing down all combinations) that the total number of combinations is 3*3*3 = 27.
That will hopefully convince you that for your example, it will be 6^6.
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one die has 6 words (i'm gonna use numbers though)

1
2
3
4
5
6

now, if you throw 2 dice, you get the following combinations:

(assume you roll 1 on the first die)
1 and 1
1 and 2
1 and 3
1 and 4
1 and 5
1 and 6

so if you roll 1 on the first die, you get the 6 possible combinations above. if you instead roll 2 on the first die you get another 6 possibilities, and another 6 when you roll 3 on the first die, and again on 4, and on 5, and on 6. did you catch it? throwing an additional die takes all the previously existing combinations, and multiplies them by 6

so while it's obvious that 1 die gives 6 possible outcomes, 2 dice will give the same 6, but MULTIPLIED by 6 (so 6^2)
three dice will give that same 6^2 but MULTIPLIED by 6, for a total of 6^3
4 dice --> 6^4
5 dice --> 6^5
6 dice (your question) --> 6^6

hope this helps!



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