There are $10$ seats in each of $10$ rows of a theatre and all the seats are numbered. What is the probablity that two friends buying tickets independently will occupy adjacent seats?

$ 
\textbf{a)}\ \dfrac{1}{55}
\qquad\textbf{b)}\ \dfrac{1}{50}
\qquad\textbf{c)}\ \dfrac{2}{55}
\qquad\textbf{d)}\ \dfrac{1}{25}
\qquad\textbf{e)}\ \text{None of above}
$
1. **Total Number of Ways to Choose Seats:**
   - There are \(10\) rows and \(10\) seats in each row, making a total of \(100\) seats.
   - The total number of ways to choose \(2\) seats out of \(100\) is given by the combination formula:
     \[
     \binom{100}{2} = \frac{100 \times 99}{2} = 4950
     \]

2. **Number of Favorable Outcomes:**
   - We need to count the number of ways in which the two friends can sit in adjacent seats.
   - In each row, there are \(9\) pairs of adjacent seats (e.g., (1,2), (2,3), ..., (9,10)).
   - Since there are \(10\) rows, the total number of adjacent pairs in the entire theatre is:
     \[
     10 \times 9 = 90
     \]

3. **Calculating the Probability:**
   - The probability \(P\) that the two friends will occupy adjacent seats is the ratio of the number of favorable outcomes to the total number of possible outcomes:
     \[
     P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{90}{4950}
     \]

4. **Simplifying the Probability:**
   - Simplify the fraction:
     \[
     \frac{90}{4950} = \frac{9}{495} = \frac{1}{55}
     \]

The final answer is \(\boxed{\frac{1}{55}}\)