Jumpy, Patient Frog Named Sally And A Math Problem About Her Life-I
A frog, namely, Sally, is jumping about the vertices of a hexagon,
A
B
C
D
E
F
, each time jumping to an adjacent vertex. In how many ways can she get from
A
to
C
in
1
2
4
3
5
6
moves?
Note: Sally is allowed to land on C during the intermediate steps.
As a little bonus, can you generalise for n jumps?
This problem is a part of my froggy, soggy set .
The answer is 2.564E+37434.
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1 solution
We want to compute 3 2 1 2 3 4 5 6 . Note that lo g 3 2 1 2 3 4 5 6 = 1 2 3 4 5 6 lo g 2 − lo g 3 = 3 7 1 6 3 . 4 8 2 … , so 3 2 1 2 3 4 5 6 = 1 0 0 . 4 8 2 … ⋅ 1 0 3 7 1 6 3 = 3 . 0 3 4 … ⋅ 1 0 3 7 1 6 3 .
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Your answer is incorrect, since you used 123456 instead of 124356
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And that's what I get for reading the problem too quickly. My mistake.
came here for this. Calculated the answer to be (2^124356-1)/3, google calculator gave me infinity for that
I approached it as the sum of all coefficients of terms of the form x 6 k + 2 and x 6 k + 4 in the expansion of ( x + x 1 ) n where n is an even integer. Well, I just love applying coefficients in everything, haha.
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How did you arrive at the generating function (x+1/x)^n? Can you please explain...
It can easily be seen that after an even number of jumps Sally can only be located at A , C or E . Let a k , c k , e k be the number of paths of length 2 k , leading from A to A , C or E respectively. Due to symmetry c k = e k . We see that these equations must hold: c k + 1 = a k + 3 c k , a k + 1 = 2 a k + 2 c k . From this, we get: c k + 2 = a k + 1 + 3 c k + 1 = 2 a k + 2 c k + 3 c k + 1 = 2 ( c k + 1 − 3 c k ) + 2 c k + 3 c k + 1 = 5 c k + 1 − 4 c k .
From the above c 0 = 0 , c 1 = 1 , we find that c k = 3 4 k − 1 (this can easily be proven by induction). Therefore, the generalised formula 3 2 n − 1 .