Random permutation of letters: The letters B, G, I, N, R are rearranged uniformly at random. What is the probability the arrangement spells BRING?

Introduction / Context:We ask for the probability that a random permutation of five distinct letters equals one specific word. This is a classic counting problem using factorials.

Given Data / Assumptions:Letters: B, R, I, N, G (5 distinct). Each permutation is equally likely.

Concept / Approach:Total permutations = 5! = 120. Exactly one of those permutations is BRING.

Step-by-Step Solution:

Verification / Alternative check:Enumerating permutations conceptually reinforces 5 positions × decreasing choices framework.

Why Other Options Are Wrong:1/54 or 1/24 do not correspond to 5! counting; the expression “5/5 x 42” is malformed.

Common Pitfalls:Using 4! accidentally (treating one letter as fixed) or miscounting distinctness.

Final Answer:1/120