Counting steps!

A person has to climb the $n^{th}$ step compulsory.But the person on first $n-1$ steps steps on it or does not step on it.Hence there are 2 possibilities for every step of the $n-1$ steps.

So by product rule, total ways $= 2^{n-1}$ .

Here $n=11$ , hence total ways $=2^{11-1}=2^{10}=\boxed{1024}$

Another way of solving this is recursively. Let's call $W(x)$ the number of ways of climbing $x$ steps. There is only one way of climbing zero steps and one way of climbing one step. So, $W(0)=W(1)=1$ .

Now, notice that if want to climb $x$ steps, we can:

1) Move 1 step up. We now have to climb $(x-1)$ steps. The number of ways is $W(x-1)$ .

2) Move 2 steps up. We now have to climb $(x-2)$ steps. The number of ways is $W(x-2)$ .

...

x) Move x steps up, and we now have no steps left to climb. The number of ways is $W(0)=1$

So, for $x$ steps ( $2<=x)$ , we have:

$W(x) = \color{#EC7300}{W(x-1)+W(x-2)+...+W(0)}$

$W(x) = \color{#EC7300}{\sum_{k=0}^{x-1}W(k)}$

Let's try to get rid of the sum:

$W(x) = W(x-1) + \color{#D61F06}{\sum_{k=0}^{x-2}W(k)}$

$W(x) = W(x-1) + \color{#D61F06}{W(x-1)}$

$W(x) = 2*W(x-1)$

Notice we keep multiplying by 2 until we get to $W(1)=1$ . So, we multiply $(x-1)$ times.

$W(x) = 2^{x-1}$

$W(11) = 2^{10} = \boxed{1024}$

Well, Nihar gave you the formula, But I would like to Explain How it works.

Here, There are 11 steps.

Now, Stepping on any of the first 10 steps is not compulsory, But it is compulsory for the 11th Step, Because it is where we have to climb to.So, There are 2 choices for the first 10 steps, But only 1 for the 11th.

Hence, By the rule of product, The number of ways to climb to the 11th Stair are:-

$2*2*2*2*2*2*2*2*2*2*1={2}^{10}=1024.$

Generalization:- For $n \ steps$ Number of ways to climb to the $nth \ step\ are\ {2}^{n-1}$

Well there is another way to think about this problem. Since for climbing $11$ steps he has to choose $10$ steps. He can therefore choose 0 or 1 or 2 or 3 .... and so so on till 10 or (n-1). So the sum is ${n-1\choose 0}$ + ${n-1\choose 1}$ + ${n-1\choose 2}$ ... ${n-1\choose n-1}$ . This sums up to $2^{n-1}$ . This is a much more difficult solution though.