algorithm - Counting ways to climb n steps with 1, 2, or 3 steps taken

Question

In a book I encountered following question:

Given N step stair, in how many number of ways can you climb if you use either 1, 2 or 3 steps at a time?

Following is the code that book has given:

 int countWays(int n){

    if(n<0)
        return 0;

    if(n == 0)
        return 1;

    else return countWays(n-1) + countWays(n-2) + countWays(n-3);
  }

I have the following concerns in understanding this code:

1. I do not understand why 1 is being returned for n=0. If there are 0 steps then obviously we do not have to climb any and 0 should be returned.

2. For n=3 function returns 4 but i can see only 3 cases i.e. (1,1,1), (1,2), (3).

Answer 1

I do not understand why 1 is being returned for n=0. If there are 0 steps then obviously we do not have to climb any and 0 should be returned.

When there are no steps you just go through without climbing, which is the one and only one way. As is pointed out in one of the comments, it can be represented as ().

For n=3 function returns 4 but i can see only 3 cases i.e. (1,1,1), (1,2), (3).

There are actually 4 cases: (1,1,1), (1,2), (2,1), (3).