Rabbits on stairs

Let us think of the problem in terms of a recurrence relation. Let $a_n$ denote the number of possible ways this can be done in a staircase having $n$ steps.

$\textbf{1.Finding base cases:}$

Now let us find the base cases.Clearly, $a_1 =1$ .

If there are $2$ steps then either it can take one two step jump or two one step jumps.So $a_2 = 2$ .

Similarly, $a_3 = 4$ .

$\textbf{2.Obtaining a recursive definition:}$

Now let us consider $n$ steps ,first if the rabbit takes a $1$ step jump then the whole process can be executed in $a_{n-1}$ steps.If it takes a $2$ step jump the process can be completed in $a_{n-2}$ steps and for a $3$ step jump in $a_{n-3}$ steps.

So writing the larger case in terms of smaller ones we get the recurrence relation $a_n = a_{n-1}+a_{n-2}+a_{n-3}$ .

Solving using the base cases we can easily find that $a_8 = \boxed{81}$ .

Denote a combination by (xyz), x, y, z are number of one , two, three step jump. It is easy to get some combinations of jumps to get the top: (012) get: 3!/2! ways to get the top (040) get: 1 ways, (121) get: 12 ways, (202): 6, (230): 10, (311): 20, (420): 15, (501): 6, (610): 7, (800): 1. So the number of difference ways is: $\boxed{81}$

I used a laborious but sure way to solve the problem. Using a spreadsheet, I found out all the combinations of step to reach the top of staircase. There are 10 combinations as follows. Please note that sum of steps is 8 in all cases.

$1 1 1 1 1 1 1 1$ $1 1 1 1 1 1 2$ $1 1 1 1 1 3$ $1 1 1 1 2 2$ $1 1 1 2 3$ $1 1 2 2 2$ $1 1 3 3$ $1 2 2 3$ $2 2 2 2$ $2 3 3$

I then calculated the permutation for each combination and the results are:

$1, 7, 6, 15, 20, 10, 6, 12, 1, 3$

And the sum of all the cases:

$s = 1 + 7 + 6 + 15 + 20 + 10 + 6 + 12 + 1 + 3 = \boxed{81}$

From step 7 to go to step 8, there is 1 way

From step 6 to step 8 there are 2 ways (directly to 8 or go to 7 & then to 8)

From step 5 to step 8 is 4 ways (directly to 8 Or to 7 Or to 6 so 1+1+2 = 4)

From step 4 the rabbit can reach 7 or 6 or 5, so the number of ways from step 4 is (1+2+4)=7

Working this way to step 0 it will give us 81 ways.