In how many distinct arrangements can the letters of “VENTURE” be written? (E appears twice; remaining letters are distinct.)

Introduction / Context:
“VENTURE” contains 7 letters with E repeated twice. Use the multiset permutation formula.

Given Data / Assumptions:

• Total letters = 7; multiplicity: E × 2; others distinct.

Concept / Approach:
Number of arrangements = 7! / 2!.

Step-by-Step Solution:

Verification / Alternative check:
Only E repeats; thus a single 2! divisor is needed.

Why Other Options Are Wrong:
5040 ignores repetition; 840 and 1260 are undercounts.

Common Pitfalls:
Over- or under-dividing by factorials when multiple letters repeat (not applicable here).

Final Answer:
2520