Problem 11 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If six students, including two particular students \(A\) and \(B\), stand in a row, then the probability that \(A\) and \(B\) are separated with one student in between them is [Online May 19, 2012] (a) \(\frac{8}{15}\) (b) \(\frac{4}{15}\) (c) \(\frac{2}{15}\) (d) \(\frac{1}{15}\)

Step-by-Step Solution

Answer

The probability is \( \frac{4}{15} \).

Step 1: Calculate Total Arrangements

First, calculate the total number of ways to arrange all six students in a row. This is done using the permutation of the students: \[ 6! = 720 \] Thus, there are 720 possible arrangements.

Step 2: Find Suitable Arrangements for A and B

Consider the positions available for six students, labeled 1 through 6. To place the specific sequence where A, another student, and B are in that order (A-X-B), the positions for A and B could be: 1. A at position 1, B at position 3.2. A at position 2, B at position 4.3. A at position 3, B at position 5.4. A at position 4, B at position 6.For each scenario (1 to 4), the other four students can be arranged among the remaining four positions. Therefore for each scenario:\[ 4! = 24 \] So there are 4 scenarios * 24 arrangements each, equaling 96 suitable arrangements when A is before B.

Step 3: Account for Reverse Order of A and B

Similarly, there's a corresponding set of arrangements when B is before A, which are:1. B at position 1, A at position 3.2. B at position 2, A at position 4.3. B at position 3, A at position 5.4. B at position 4, A at position 6.Each of these scenarios also has 24 different arrangements of the remaining four students:\[ 4! = 24 \]Thus, there are again 96 suitable arrangements.

Step 4: Total Suitable Arrangements and Probability

Combine both A before B and B before A positions, we get:96 (A before B) + 96 (B before A) = 192 total suitable arrangements.The probability that A and B are separated by one student is then the quotient of the suitable arrangements over the total arrangements:\[ \frac{192}{720} = \frac{4}{15} \] So the probability is \( \frac{4}{15} \).

Key Concepts

Permutations

Permutations are a fundamental concept in probability and combinatorics, related to the different ways of arranging a set of objects. In the given problem, we are asked to consider the different possible ways in which students can be positioned in a line. When dealing with permutations, each unique arrangement is important and order matters.
For six students, this means calculating the total number of ways to arrange them in a sequence. This is done using factorials. The factorial of a number, denoted by an exclamation point, represents the product of all positive integers up to that number. So, for six students:
- The number of permutations is given by: \[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \] arrangements.
Knowing how to calculate permutations is crucial when you have to organize objects or people in specific positions. Understanding permutations helps in many real-life scenarios, such as organizing students, arranging schedules, or even solving puzzles.

Combinatorics

Combinatorics is the branch of mathematics that deals with counting, arrangement, and combination of objects. It goes hand in hand with permutations but includes more selective arrangements where order might not always matter.
In our specific problem, combinatorics is used to calculate how many suitable arrangements exist where two specific students, A and B, are separated by exactly one other student. Here, we focus on combinations that follow a distinct pattern, A-X-B or B-X-A, within a sequence of six, forming scenarios such as:

• A at position 1, B at position 3
• A at position 2, B at position 4
• And so on...

For each suitable position of A and B, the remaining students can fill the other slots. Each set of configurations for A-B and B-A account for numerous combinations of other students, leveraging the combinatorial principle of counting these possible variations. The key principle here is understanding how to pick and arrange such configurations from a larger set, vital for solving more complex tasks involving arrangement and selection.

Arrangement of Students

Arranging students involves organizing them in a particular order, which brings into play concepts from permutations and combinatorics. Specifically, this involves positioning certain individuals (students A and B) in a specific pattern within the row.
The importance of this exercise lies in learning how different positional requirements correlate to possible outcomes. By exploring scenarios where A is positioned before B, or vice versa, we realize how constraints alter the total feasible positions.
The problem splits these arrangements into two halves, such that:

• A-X-B configurations account for half.
• B-X-A configurations account for the other half.

For any given setup (like A at 1 and B at 3), the other four students have their arrangements, calculated as 24 each time. Learning this helps grasp constraints-based arrangement problems, which are common in scheduling, seating charts, and even organizing competitions.