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    Mike starts at one corner of a tetrahedron. In a move, he walks along any edge to a different corner. In how many ways can he end up where he started after 6 moves?
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    (Original post by FXLander)
    Mike starts at one corner of a tetrahedron. In a move, he walks along any edge to a different corner. In how many ways can he end up where he started after 6 moves?
    First I'd draw a tetrahedron and pick a random corner with no loss of generality since it can be rotated.

    We can see that you can get from any one corner to another. Remember to get to the same vertex on the sixth move, after the fifth move, we cannot be on the required vertex.

    Solution (I hope it's correct):
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    We may be at any vertex (apart from the start vertex) after the 5th move to access the start vertex, down any of the three possible edges. In the first move, we may move along any of the three edges. In the remaining moves, we may also travel along one of the three edges to a different vertex each time. But this will over count the moves when we are at the start vertex after the fifth move, since we cannot 'stay' on a move. Let V_n be the number of ways of getting back to the vertex in n moves. Then V_n = 3^{n-1}-V_{n-1}. And we know that V_1=0. So

    V_6=3^5-(3^4-3^3+3^2-3^1)=3^5-60=183.

    We may generalise this for n moves that
    V_n=3^{n-1}-3^{n-2}+\dots+(-1)^{n}3=\frac{3^{n-1}(1-(-1/3)^{n-1})}{1+1/3}=\frac{3}{4}(3^{n-1}+(-1)^n).

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    How did you set up that recurrence relation? Why is V_n = 3^n-1 - V_n-1?
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    (Original post by FXLander)
    How did you set up that recurrence relation? Why is V_n = 3^n-1 - V_n-1?
    The only way you can't get to the start vertex in 1 move is if you are already at the start vertex. And conversely, if you're not at the start vertex, there's exactly one way of getting to the start in one move.

    So, # of routes of getting to start in n moves = (number of routes of length n-1 that don't end at the start point)
    = (number of routes of length n-1) - (number of routes of length n-1 that end at the start point)
    = (number of routes of length n-1) - V_{n-1}

    For the total number of routes of length k, note that at each move, you have 3 choices of possible move.
 
 
 
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