+422 
+422 Here you go!
Let be a heptagon. A frog starts jumping at vertex . From any vertex of the heptagon except , the frog may jump to either of the two adjacent vertices. When it reaches vertex , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than jumps that end at .
Solution[edit]
This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line . We can count the number of left/right (L/R) paths of length that start at and end at either or .
We can visualize the paths using the common grid counting method by starting at the origin , so that a right (R) move corresponds to moving 1 in the positive direction, and a left (L) move corresponds to moving 1 in the positive direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines or . Letting be the number of such paths from to under these constraints, we have the following base cases:
and recursive step for .
The filled in grid will look something like this, where the lower-left corresponds to the origin:
The bolded numbers on the top diagonal represent the number of paths from to in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from to in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line . The total number of ways is 351.
The link is ---> here
That is where i found the answer!
Ha! Really smart for one! Unlike me....