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

Difficulty: Easy

Correct Answer: 1/120

Explanation:


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:

Total outcomes = 120Favorable outcomes = 1Probability = 1/120


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

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