Animated Solution for Mathematics - Probability: Comprehension Passage

Imagine you are standing in a music class, the air filled with the anticipation of an examination. Five students, $S_1,S_2,S_3,S_4,S_5$, are waiting to be seated in five chairs, $R_1,R_2,R_3,R_4,R_5$.

Initially, everything is orderly: $S_i$ sits in $R_i$. But life, much like a JEE Advanced problem, rarely stays orderly for long.

On exam day, the seats are shuffled. We are tasked with finding the probability that $S_1$ gets their original seat $R_1$, while everyone else is forced into a seat that is NOT their own.

The Total Sample Space

Before we dive into the constraints, we must understand the universe of possibilities. We have 5 distinct students and 5 distinct seats.

The number of ways to arrange 5 distinct items in 5 distinct positions is given by $5!$. Calculating this, we get:

This is our denominator, the total sample space. It represents every possible seating chart, from the perfectly ordered to the completely chaotic.

The Constraint of $S_1$

Now, let us apply the first condition. The problem demands that $S_1$ sits in $R_1$.

This is a powerful constraint. By fixing $S_1$ in $R_1$, we remove them from the pool of uncertainty. There is only $1$ way to do this.

We are left with 4 students ($S_2,S_3,S_4,S_5$) and 4 seats ($R_2,R_3,R_4,R_5$). The problem has effectively shrunk; we are now looking at a 4-student problem with a very specific, restrictive rule.

The Concept of Derangement

Here is where the magic happens. The problem states that NONE of the remaining students can sit in their originally allotted seat.

This means $S_2$ cannot sit in $R_2$, $S_3$ cannot sit in $R_3$, and so on. In combinatorics, a permutation where no element appears in its original position is called a Derangement.

We denote the number of derangements of $n$ items as $D_n$. We need to find $D_4$. The formula for derangement is derived from the Principle of Inclusion-Exclusion:

The Calculation

We need to calculate $D_4$ using the following expansion:

We know $4!=24$. Substituting the factorial values, the expression becomes:

The $1$ and $-1$ cancel out beautifully, leaving us with:

To solve this, we find a common denominator of $24$:

There are $9$ ways for the remaining 4 students to be seated such that none of them are in their original seats.

The Final Probability

We have our favorable outcomes ($9$) and our total outcomes ($120$). The probability $P(E)$ is simply:

Simplifying this fraction by dividing both numerator and denominator by $3$, we get:

This is the elegance of mathematics. We started with a complex scenario of 120 possibilities, applied constraints to reduce the chaos, identified the underlying structure of derangements, and arrived at a clean, simple fraction.