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)`.
Number of ordered pairs `(i,j)` where `i ne j` for which `P(i,100-i)=P(i,100-j)` is

Options

A. (a) `50`
B. (b) `99`
C. (c) `100`
D. (d) `101`

Explanation

To solve the problem, we need to find the number of ordered pairs \((i, j)\) such that \(i \neq j\) and \(P(i, 100-i) = P(j, 100-j)\).

### Step-by-Step Solution:

1. **Understanding \(P(a, b)\)**:
The number of paths from \((0, 0)\) to \((a, b)\) is given by:
\[ P(a, b) = \frac{(a+b)!}{a!b!} \]
This is derived from the fact that to reach \((a, b)\), we need to make \(a\) rightward moves and \(b\) upward moves, which can be arranged in \((a+b)\) total moves.

2. **Setting Up the Equation**:
We need to find pairs \((i, j)\) such that:
\[ P(i, 100-i) = P(j, 100-j) \]
This translates to:
\[ \frac{(i + (100-i))!}{i!(100-i)!} = \frac{(j + (100-j))!}{j!(100-j)!} \]
Simplifying this gives:
\[ \frac{100!}{i!(100-i)!} = \frac{100!}{j!(100-j)!} \]
Thus, we have:
\[ \frac{1}{i!(100-i)!} = \frac{1}{j!(100-j)!} \]
This implies:
\[ j!(100-j)! = i!(100-i)! \]

3. **Using the Properties of Binomial Coefficients**:
From the properties of binomial coefficients, we know:
\[ \binom{100}{i} = \binom{100}{j} \]
This holds true if:
- \(i = j\) or
- \(i + j = 100\)

Since we are interested in the case where \(i \neq j\), we focus on the second condition:
\[ i + j = 100 \]

4. **Finding Valid Pairs**:
The pairs \((i, j)\) can be expressed as:
\[ (i, 100-i) \]
where \(i\) can take values from \(0\) to \(100\). However, since \(i\) cannot equal \(j\), we exclude the case where \(i = 50\) (which gives \(j = 50\)).

5. **Counting the Ordered Pairs**:
The values of \(i\) can range from \(0\) to \(100\) (inclusive), giving us \(101\) possible values. Excluding the case where \(i = 50\), we have:
\[ 101 - 1 = 100 \]
Thus, the total number of ordered pairs \((i, j)\) such that \(i \neq j\) is \(100\).

### Final Answer:
The number of ordered pairs \((i, j)\) where \(i \neq j\) for which \(P(i, 100-i) = P(j, 100-j)\) is \(100\).