A path of length `n` is a sequence of points `(x_(1),y_(1))`, `(x_(2),y_(2))`,….,`(x_(n),y_(n))` with integer coordinates such that for all `i` between `1` and `n-1` both inclusive,
either `x_(i+1)=x_(i)+1 ` and `y_(i+1)=y_(i)` (in which case we say the `i^(th)` step is rightward)
or `x_(i+1)=x_(i)` and `y_(i+1)=y_(i)+1` ( in which case we say that the `i^(th)` step is upward ).
This path is said to start at `(x_(1),y_(1))` and end at `(x_(n),y_(n))`. Let `P(a,b)`, for `a` and `b` non-negative integers, denotes the number of paths that start at `(0,0)` and end at `(a,b)`.
The value of `sum_(i=0)^(10)P(i,10-i)` is

Options

A. (a) `1024`
B. (b) `512`
C. (c) `256`
D. (d) `128`

Explanation

To solve the problem, we need to find the value of the summation \( \sum_{i=0}^{10} P(i, 10-i) \), where \( P(a, b) \) denotes the number of paths from \( (0,0) \) to \( (a,b) \) using rightward and upward steps.

### Step-by-Step Solution:

1. **Understanding \( P(a, b) \)**:
- The number of paths \( P(a, b) \) can be calculated using the formula:
\[ P(a, b) = \frac{(a+b)!}{a! \, b!} \]
- This formula arises because to reach the point \( (a, b) \), we need to make \( a \) rightward moves and \( b \) upward moves, totaling \( a + b \) moves. The number of ways to arrange these moves is given by the factorial division.

2. **Setting up the summation**:
- We need to compute:
\[ \sum_{i=0}^{10} P(i, 10-i) \]
- Substituting the formula for \( P(i, 10-i) \):
\[ \sum_{i=0}^{10} P(i, 10-i) = \sum_{i=0}^{10} \frac{(i + (10-i))!}{i! \, (10-i)!} = \sum_{i=0}^{10} \frac{10!}{i! \, (10-i)!} \]

3. **Recognizing the binomial coefficient**:
- The expression \( \frac{10!}{i! \, (10-i)!} \) is the binomial coefficient \( \binom{10}{i} \).
- Thus, we rewrite the summation:
\[ \sum_{i=0}^{10} P(i, 10-i) = \sum_{i=0}^{10} \binom{10}{i} \]

4. **Using the binomial theorem**:
- According to the binomial theorem, the sum of the binomial coefficients for a fixed \( n \) is given by:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
- For \( n = 10 \):
\[ \sum_{i=0}^{10} \binom{10}{i} = 2^{10} = 1024 \]

5. **Final answer**:
- Therefore, the value of \( \sum_{i=0}^{10} P(i, 10-i) \) is:
\[ \boxed{1024} \]