Problem 49 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

There are \(n\) persons \((n \geq 3)\), among whom are \(A\) and \(B\), who are made to stand in a row in random order. Probability that there is exactly one person between \(A\) and \(B\) is (A) \(\frac{n-2}{n(n-1)}\) (B) \(\frac{2(n-2)}{n(n-1)}\) (C) \(2 / n\) (D) none os these

Step-by-Step Solution

Answer

The probability is option (B): \(\frac{2(n-2)}{n(n-1)}\).

Step 1: Calculate Total Arrangements

First, we calculate the total number of ways to arrange the \(n\) persons in a row. Since there are \(n\) unique persons, the number of arrangements is given by \(n!\).

Step 2: Identify Arrangements for A and B

Next, consider the specific cases where there is one person between \(A\) and \(B\). Two configurations are possible: \(A\) _ \(B\) or \(B\) _ \(A\). Thus, the positions for \(A\) can be from 1 to \(n-2\) (to leave a space for the person in between and for \(B\) on one side), providing \((n-2)\) valid positions for each configuration.

Step 3: Calculate Arrangements with Exactly One Person Between

For each valid position of \(A\), we can place \(B\) such that there is one person between them, giving \((n-2)\) positions as calculated. Since \(A\) and \(B\) can switch places, we multiply by 2. For each configuration (A _ B or B _ A), there are \((n-2)\) possible people to place between them, making this count \(2(n-2)\).

Step 4: Calculate Probability

The probability is obtained by dividing the number of favorable arrangements by the total arrangements. Thus, we have:\[\text{Probability} = \frac{2(n-2)}{n!}\] However, since \(A\) and \(B\) are fixed first, there are \(2(n-2)\) specific arrangements where exactly one person is between them, for which we adjust to possible ways to permute the remaining persons, simplifying the expression:\[\text{Probability} = \frac{2(n-2)}{n(n-1)}\]

Step 5: Conclusion

Matching the probability \(\frac{2(n-2)}{n(n-1)}\) with the available options, we see it matches option B.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and combinations of objects. It is the foundation of many probability problems, including permutation and combination issues. When tackling problems of this nature, it's essential to understand some basic principles:

• Factorial Notation: The symbol \(!\) denotes a factorial, which is the product of an integer and all the integers below it. For example, \(n! = n \times (n-1) \times (n-2) \times \, ... \, \times 1\).
• Permutation: This involves arranging objects in a specific order. The number of ways to arrange \(n\) objects is \(n!\), and it becomes particularly important in permutation problems.

In this particular exercise, combinatorics helps us calculate the total number of ways to arrange people and the specific arrangements where certain conditions (like keeping a person between two others) are met. By mastering these concepts, students can explore more complex scenarios like probability distributions and conditional probabilities.

Permutation Problems

Permutation problems focus on the arrangement of a specific set of elements. In scenarios like the exercise provided, you're often concerned with arranging people or items in a sequence with specific constraints.

Two key insights in permutation problems include:

• Fixed Positioning: Sometimes, you might want certain elements to always be neighbors or separated by a certain number of elements. For instance, placing exactly one person between two specified people, like \(A\) and \(B\), is a typical constraint.
• Switching Arrangements: When dealing with two interchangeable elements, remember to multiply by 2 to account for both arrangements (e.g., the sequence can start with \(A\) or \(B\)).

Applying these considerations to our problem, understanding how to position \(A\) and \(B\) with exactly one person in between helps students grasp the dynamic nature of permutation problems. Grasping these fundamentals is crucial for solving advanced problems in combinatorics and probability.

Probability Theory

Probability theory is a field in mathematics that involves calculating the likelihood of different outcomes. It's a key concept in many mathematical problems, providing a quantitative description of the likelihood that specific events will occur. Let's dive into its application:

• Total Possible Outcomes: This refers to all the ways an event can happen. In the exercise, this is represented by \(n!\), the total arrangements possible for \(n\) people.
• Favorable Outcomes: These are the outcomes that fulfill the conditions you're interested in, like having exactly one person standing between \(A\) and \(B\).
• Calculating Probability: To find the probability, you take the ratio of favorable outcomes to the total possible outcomes. This ensures you measure how likely an event is in the most comprehensive way possible.

In our exercise, applying probability theory helps us determine the chances of having exactly one person between two others in a random line-up. Mastery of probability theory allows students to analyze real-world situations and quantify the likelihood of various outcomes.