In an examination, 5 students have been allotted their seats as per their roll numbers. The number o

Q: Wait, why can't we just use simple subtraction? Can't we just do $5! - 5$?

That is a common trap! Simply subtracting the cases where one student is in the correct seat would lead to massive overcounting because you would be subtracting the cases where two or more students are in their correct seats multiple times. The Inclusion-Exclusion Principle is the only way to correctly account for these overlaps.

Q: Is there a recursive way to solve this?

Yes! Derangements follow a beautiful recurrence relation: $D_n = (n-1)(D_{n-1} + D_{n-2})$. This is incredibly powerful for larger $n$ because it allows you to build up from $D_1=0$ and $D_2=1$ without needing the full factorial expansion.

Q: Why does the formula start with $1 - 1/1!$?

This comes from the expansion of $e^{-1}$. The derangement formula is essentially the integer closest to $n!/e$. The alternating series $1 - 1/1! + 1/2! - 1/3! + \dots$ is the Taylor series expansion of $e^x$ evaluated at $x=-1$. It is a beautiful connection between combinatorics and calculus!

Question

In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _____

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem of arranging five students ($S_1, S_2, S_3, S_4, S_5$) into five seats ($C_1, C_2, C_3, C_4, C_5$) such that no student sits in their own assigned seat is a classic combinatorial challenge known as a **Derangement**.

In mathematical terms, we are looking for the number of permutations of the set $\{1, 2, 3, 4, 5\}$ such that no element $i$ remains in its original position $i$. This constraint defines a permutation without any "fixed points."

### The Inclusion-Exclusion Principle

To solve this, we employ the **Inclusion-Exclusion Principle**. We start with the total number of permutations, which is $5! = 120$, and systematically remove arrangements where at least one student is in their correct seat.

The general formula for the number of derangements $D_n$ is derived as follows:

$$D_n = n! - \binom{n}{1}(n-1)! + \binom{n}{2}(n-2)! - \binom{n}{3}(n-3)! + \dots + (-1)^n \binom{n}{n}(0!)$$

This expression simplifies to the elegant summation:

$$D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$$

### Final Calculation for $n=5$

Applying the formula for $n=5$, we calculate:

$$D_5 = 5! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right)$$

Since $1 - 1/1! = 0$, the expression simplifies to:

$$D_5 = 5! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right)$$

Expanding this, we get:

$$D_5 = \frac{5!}{2!} - \frac{5!}{3!} + \frac{5!}{4!} - \frac{5!}{5!} = 60 - 20 + 5 - 1 = 44$$

The total number of ways to arrange the students such that no one sits in their own seat is **44**.

### The Recursive Perspective

Alternatively, we can use the recursive relation $D_n = (n-1)(D_{n-1} + D_{n-2})$. This method is often more efficient for manual calculation.

Starting with the base cases $D_1 = 0$ and $D_2 = 1$, we compute the sequence:

$$D_3 = (3-1)(D_2 + D_1) = 2(1 + 0) = 2$$

$$D_4 = (4-1)(D_3 + D_2) = 3(2 + 1) = 9$$

$$D_5 = (5-1)(D_4 + D_3) = 4(9 + 2) = 4(11) = 44$$

Both the Inclusion-Exclusion Principle and the recursive relation yield the same result, confirming the robustness of our solution. Embracing these multiple perspectives is a vital skill for success in your JEE preparation.

Animated Solution for Mathematics - Permutations and Combinations: In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _____

Enter Numerical Value:

Visualized Solution

Visualizing the Scenario

The Core Condition

Defining Derangement

The Derangement Formula

Substituting $n = 5$

Simplifying the First Terms

Expanding Factorials

Calculating the First Term

Calculating the Second Term

Calculating the Remaining Terms

Final Arithmetic

Key Takeaway

The Sigma Insight: Derangement Principle

Analyzing the Setup

The Inclusion-Exclusion Principle

Final Calculation for $n = 5$

The Recursive Perspective

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