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In the Tower of Hanoi problem, suppose you have three pegs and n disks. How many different arrangements are there of the n disks on three pegs so that no disk is on top of a smaller disk?

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Someone correct me if I'm doing this completely wrong. I believe the trick is to start with the bottom disk. You have 3 places to put it. Now the next largest disk, you still have 3 places to put this. Continue on, and each disk you will have exactly 3 choices of where to put it. Thus, there should be \[3^n\]arrangements.
And since larger disks must be below smaller disks, in every possible setting of the towers of hanoi, you can place the largest disk first, then the second largest, and so on.
isn't it 2^n - 1 ?? I remember doing 2^n-1 ... though i never thought behind it??

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That's the minimum number of steps it takes to solve, not the different arrangements.
okay thank you.. i'm was thinking wrong, my first intuition was that the first peg can have n disks, the second, another n, and so continuing on this fashion i thought it was n^3. looks like i thought the other way around, I always get confused on these kinds of problems..
i was*
I still have that problem sometimes :/
haha but I think you got this one right though, thanks. it's like the question of how many functions from a set of m elements to n elements are there. haha

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