We call a path Valid if
i. It only comprises of the following kind of steps:
A. $(x, y) \rightarrow (x + 1, y + 1)$
B. $(x, y) \rightarrow (x + 1, y - 1)$
ii. It never goes below the x-axis.
Let $M(n)$  = set of all valid paths from $(0,0) $, to $(2n,0)$, where $n$ is a natural number. 
Consider a Valid path $T \in M(n)$. 
Denote $\phi(T) = \prod_{i=1}^{2n} \mu_i$,
where $\mu_i$= 
a) $1$, if the $i^{th}$ step is $(x, y) \rightarrow (x + 1, y + 1)$
b) $y$,  if the $i^{th} $ step is $(x, y) \rightarrow (x + 1, y - 1)$ 
Now Let $f(n) =\sum _{T \in M(n)} \phi(T)$. Evaluate the number of zeroes at the end in the decimal expansion of $f(2021)$
1. **Define the problem and the function $f(n)$:**
   - We need to evaluate the number of zeroes at the end in the decimal expansion of $f(2021)$.
   - $f(n) = \sum_{T \in M(n)} \phi(T)$, where $M(n)$ is the set of all valid paths from $(0,0)$ to $(2n,0)$.
   - A valid path consists of steps $(x, y) \rightarrow (x + 1, y + 1)$ or $(x, y) \rightarrow (x + 1, y - 1)$ and never goes below the x-axis.
   - $\phi(T) = \prod_{i=1}^{2n} \mu_i$, where $\mu_i = 1$ if the $i^{th}$ step is $(x, y) \rightarrow (x + 1, y + 1)$ and $\mu_i = y$ if the $i^{th}$ step is $(x, y) \rightarrow (x + 1, y - 1)$.

2. **Analyze the parity of $f(n)$:**
   - We need to show that $f(n)$ is odd for all $n \in \mathbb{N}$.
   - Consider the steps as moving either in the North East (NE) direction or in the South East (SE) direction.
   - Reverse engineer the path starting from $(2n,0)$ and moving towards the left.
   - If the path touches $y=2$ at any point, $\phi(T)$ becomes even because $\mu_i = y$ for some $i$ will be 2.
   - To avoid touching $y=2$, the path must vary between $y=0$ and $y=1$ only.

3. **Construct the valid path:**
   - Since $2n$ is even, we can construct a valid path by alternating between going up and down, i.e., $(x, y) \rightarrow (x + 1, y + 1)$ followed by $(x, y) \rightarrow (x + 1, y - 1)$.
   - This ensures that the path never goes below the x-axis and stays between $y=0$ and $y=1$.

4. **Calculate $\phi(T)$ for the valid path:**
   - For the valid path that alternates between $y=0$ and $y=1$, $\phi(T)$ will be the product of 1's and 0's.
   - Specifically, $\phi(T) = 1 \cdot 1 \cdot 1 \cdot \ldots \cdot 1 = 1$ for all steps $(x, y) \rightarrow (x + 1, y + 1)$.
   - Since the path does not touch $y=2$, $\phi(T)$ remains 1 for all valid paths.

5. **Determine the parity of $f(n)$:**
   - Since $\phi(T) = 1$ for all valid paths $T \in M(n)$, $f(n)$ is the sum of 1's.
   - The number of valid paths is odd, so $f(n)$ is odd.

6. **Conclusion:**
   - Since $f(n)$ is odd for all $n \in \mathbb{N}$, $f(2021)$ is also odd.
   - An odd number has no zeroes at the end of its decimal expansion.

The final answer is $\boxed{0}$