There are four balls of different colours and four boxes of colours, same as those of the balls. The

Q: Wait, why can't I just say the first ball has 3 choices, the second has 3, and so on? Why is the answer not just $3 \times 3 \times 2 \times 1$?

That is a classic trap! The number of choices for the second ball depends on what the first ball chose. If the first ball took the box that the second ball *wanted*, the second ball's options change. This dependency means simple multiplication fails. We need the Inclusion-Exclusion Principle to account for these dependencies correctly.

Q: Is there a recursive way to solve this without the big formula?

Absolutely! Derangements follow a beautiful recurrence relation: $D_n = (n-1)(D_{n-1} + D_{n-2})$. For $n=4$, you can find $D_4$ if you know $D_3$ and $D_2$. It is a great way to verify your answer during the exam and often faster than the summation formula.

Q: Why do we use $0!$ in the formula? Doesn't that seem weird?

It feels counterintuitive, but mathematically, $0!$ is defined as $1$. This definition is essential for the consistency of the Taylor series and combinatorial formulas like this one. Without $0! = 1$, the Inclusion-Exclusion formula would break! Always trust the definition, even when it feels strange.

Question

There are four balls of different colours and four boxes of colours, same as those of the balls. The number of ways in which the balls, one each in a box, could be placed such that a ball does not go to a box of its own colour is .........

Sigma Percentile · Accepted Answer

### The Art of Perfect Mismatch: Mastering Derangements

Welcome, future engineer! Today, we are diving into a problem that sounds deceptively simple but hides a profound combinatorial truth.

Imagine you are standing in a room with four colored balls—Red, Blue, Green, and Orange—and four corresponding boxes. Your goal is to place exactly one ball in each box, but with a twist: no ball is allowed to rest in the box of its own color.

This is not just a puzzle; it is the gateway to understanding the **Derangement**—a fundamental concept in probability and combinatorics that frequently appears in JEE Advanced.

### Phase 1: Visualizing the Constraint

Why is this hard? If we had no constraints, the number of ways to arrange four balls in four boxes would simply be $4! = 24$.

But the constraint—that no ball can be in its own box—acts like a filter. It removes the 'identity' permutation (where everything is correct) and many other 'partially correct' permutations.

We are looking for the total number of permutations where the set of fixed points is empty. In mathematical terms, we are looking for the number of derangements, denoted as $D_n$. For $n=4$, we are hunting for $D_4$.

### Phase 2: The Inclusion-Exclusion Principle

To solve this, we use the Principle of Inclusion-Exclusion. We start with the total number of permutations ($4!$) and subtract the cases where at least one ball is in the correct box.

But wait! If we subtract cases where the Red ball is in the Red box, we might have over-subtracted cases where both the Red and Blue balls are in their correct boxes. We must add those back, then subtract the cases where three are correct, and so on.

This leads us to the elegant formula:

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

This formula is the heartbeat of the problem. It accounts for all the overlaps and dependencies that make simple multiplication impossible.

### Phase 3: The Calculation

Let us apply this to our specific case, $n=4$. Substituting into our formula, we get:

$$D_4 = 4! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)$$

We know that $0! = 1$ and $1! = 1$. Therefore, the first two terms are $\frac{1}{1} - \frac{1}{1} = 0$.

This is a beautiful, universal property of derangements: the first two terms always cancel out! We are left with:

$$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$$

Distributing the $24$ into the parentheses makes the arithmetic trivial:

$$D_4 = \frac{24}{2} - \frac{24}{6} + \frac{24}{24}$$

$$D_4 = 12 - 4 + 1$$

$$D_4 = 9$$

### The Takeaway

There are exactly **9** ways to arrange these balls such that none are in their matching box. It is a small, elegant number, but the journey to get there taught us about the power of the Inclusion-Exclusion Principle.

As a pro-tip for your JEE preparation, do not just memorize the formula; memorize the sequence of derangements: $D_1=0, D_2=1, D_3=2, D_4=9, D_5=44$.

Having these at your fingertips will save you precious seconds in the exam hall. Keep practicing, keep visualizing, and remember: even when everything seems to be in the wrong place, there is a beautiful mathematical order governing the chaos.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Constraint

Defining Derangement $D_{n}$

Substitution for $n = 4$

Simplifying the First Terms

Expanding the Factorials

Distributing the Multiplier

Final Calculation

Pro-Tip for JEE

The Sigma Insight: Derangement Principle

The Art of Perfect Mismatch

Phase 1

Phase 2

Phase 3

The Takeaway

Similar Questions

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 _____