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Question #40571

Difficulty: 8/10 | 🟢 Free

The Question

Six identical red balls and eight distinct blue balls are arranged randomly in a circle. Let $P$ be the probability that no two red balls are adjacent. Then the value of $P$ is:

Options

A.
$\frac{7}{429}$
B.
$\frac{14}{143}$
C.
$\frac{4}{143}$
D.
$\frac{56}{143}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Gap Method (Non-adjacency)|Combinatorial Probability|Axiomatic Probability Constraints|Exhaustive Events Concept

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Gap Method in circular permutations alongside the axiomatic constraints of indistinguishable items.

A common lethal mistake (the trap) is placing the 6 identical red balls first (1 way), creating 6 gaps, and then distributing the 8 distinct blue balls into these gaps ( $\binom{7}{5} \times 8!$ ways).

Students fail to realize that the 6 gaps created by identical items in a circle are initially indistinguishable and possess a rotational symmetry of order 6.

Placing distinct items into them overcounts the favorable cases by exactly a factor of 6, leading to the trap answer 14/143.

The correct approach places the distinct items first to create distinguishable gaps.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Gap Method (Non-adjacency)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free