Given:
- Lines \( L_{2n-1} \) (\( n = 1, 2, \dots, 10 \)) are parallel to each other.
- Lines \( L_{2n} \) (\( n = 1, 2, \dots, 10 \)) pass through a common point \( P \).
Step 1: Points of Intersection between \( L_{2n-1} \) and \( L_{2m} \)
Since all \( L_{2n-1} \) lines are parallel, they do not intersect among themselves. Similarly, all \( L_{2n} \) lines pass through the point \( P \), so they intersect at \( P \) and do not form additional intersection points among themselves.
However, each line \( L_{2n-1} \) intersects each line \( L_{2m} \) exactly once (since they are not parallel), leading to:
\[ 10 \times 10 = 100 \text{ intersection points} \]
Step 2: Points of Intersection among \( L_{2n} \) Lines
All \( L_{2n} \) lines pass through the common point \( P \). Therefore, there is exactly one intersection point among these lines at \( P \).
Step 3: Total Number of Points of Intersection
The total number of points of intersection is given by:
\[ 100 + 1 = 101 \]
Conclusion: The maximum number of points of intersection of pairs of lines from the set \( \{L_1, L_2, \dots, L_{20}\} \) is 101.
Let \( A = \{-3, -2, -1, 0, 1, 2, 3\} \). A relation \( R \) is defined such that \( xRy \) if \( y = \max(x, 1) \). The number of elements required to make it reflexive is \( l \), the number of elements required to make it symmetric is \( m \), and the number of elements in the relation \( R \) is \( n \). Then the value of \( l + m + n \) is equal to:
For hydrogen-like species, which of the following graphs provides the most appropriate representation of \( E \) vs \( Z \) plot for a constant \( n \)?
[E : Energy of the stationary state, Z : atomic number, n = principal quantum number]
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: