Let's denote $F(N)$ as the number of ways to reach the $N^{\text{th}}$ stair.
- For $N = 1,$ there is only one way$: F(1) = 1.$
- For $N = 2,$ there are two ways$: F(2) = 2. \;($She can either take two $1-$ step moves or one $2-$ step move$.)$
- For $N = 3,$ there are three ways$: F(3) = 3.\; ($She can either take three $1-$ step moves, one $2-$ step move followed by one $1-$ step move, or one $1-$ step move followed by one $2-$ step move$.)$
- For $N = 4,$ there are five ways$: F(4) = 5. \;($She can either take four $1-$ step moves, one $2-$ step move followed by two $1-$ step moves, two $1-$ step moves followed by one $2-$ step move, one $1-$ step moves followed by one $2-$ step move followed by one 1-step or one $2-$ step move followed by one $2-$ step move$.)$
Now, let's calculate $F(5)$ using the previous values:
- $F(5) = F(4) + F(3)$
- $F(5) = 5 + 3$
- $F(5) = 8$
So, the value of $F(5)$ is $8.$
Correct Answer: A
$\textbf{PS:}$ Here are the $8$ ways for Ankita to reach the $5^{\text{th}}$ stair:
- $1 + 1 + 1 + 1 + 1$
- $1 + 1 + 1 + 2$
- $1 + 1 + 2 + 1$
- $1 + 2 + 1 + 1$
- $2 + 1 + 1 + 1$
- $1 + 1 + 2 + 2$
- $1 + 2 + 2$
- $2 + 1 + 2$
These are the $8$ distinct ways Ankita can reach the $5^{\text{th}}$ stair.
$\textbf{Short Method:}$ We can calculate $F(5)$ using the Fibonacci sequence:
- $F(1) = 1 (1$ way to reach the $1$st stair$).$
- $F(2) = 2 (2$ ways to reach the $2$nd stair$: 1+1 \;\text{or}\; 2).$
- $F(3) = 3 (3$ ways to reach the $3$rd stair$: 1+1+1, 1+2, \text{or}\; 2+1).$
Now, for each subsequent step, $F(N) = F(N-1) + F(N-2).$
Calculate $F(4)$ and $F(5)$ using recurrence relation:
- $F(4) = F(3) + F(2) = 3 + 2 = 5$
- $F(5) = F(4) + F(3) = 5 + 3 = 8$