A rabbit is sitting at the base of a staircase which has 10 steps. It proceeds to the top of the staircase by climbing either one step at a time or two steps at a time. The number of ways it can reach the top is

Let the rabbit take $$a$$ times one step and $$b$$ times two steps.

So, $$a+2b=10$$

The possible values of $$\left(a,b\right)$$ can be $$\left(0,5\right),\left(2,4\right),\left(4,3\right),\left(6,2\right),\left(8,1\right),\left(10,0\right)$$

Now, when $$a=0$$, $$b=5$$, the person basically is taking all two steps at a time, 5 times.

So, number of ways of doing this = $$\dfrac{5!}{5!}$$

Similarly, when $$a=2$$, $$b=4$$, the person basically is taking 2 times one step at a time and 4 times two step at a time.

So, number of ways of doing this = $$\dfrac{6!}{4!2!}$$

So, overall total number of ways to reach the top

=$$\dfrac{5!}{5!}+\dfrac{6!}{4!2!}+\dfrac{7!}{4!3!}+\dfrac{8!}{6!2!}+\dfrac{9!}{8!1!}+\dfrac{10!}{10!}$$

=$$1+15+35+28+9+1=89$$ ways