Q.If L is a list with 6 elements, then which of the following statements will raise an exception ? (A) L.pop(1) (B) L.pop(6) (C) L.insert(1,6) (D) L.insert(6,1)

The key idea is Python list indexing and how pop() and insert() treat an out-of-range index differently.

A list L with 6 elements has valid indices 0 to 5.

• L.pop(index) needs an existing index (0 <= index < length). L.pop(6) targets index 6, which does not exist, so it raises an IndexError.
• L.pop(1) is valid — index 1 exists. …

Python lists are 0-indexed, and the key is how different list methods treat an index at the boundary.

For a list L with 6 elements, the valid indices for accessing or removing an existing element are 0, 1, 2, 3, 4, 5. Any attempt to remove an element at an index outside this range raises an IndexError.

The insert() method behaves differently: it says where to place a new element. If the index equals the list length, the element is appended; if the index is even larger, it is still appended. So insert() is more forgiving than pop().

Take a sample list L = [10, 20, 30, 40, 50, 60] where len(L) is 6.

1. (A) L.pop(1) — index 1 exists, so this removes 20. No exception.

2. (B) L.pop(6) — pop() needs 0 <= index < len(L). Here len(L) is 6, so the valid indices are 0 to 5. Index 6 is out of range, so this raises an IndexError.

   Watch out

   A common mistake is confusing the length of the list with the highest valid index. For a list of length N, the highest valid index is N minus 1, not N. …