There are four balls of different...

MathematicsDerangement PrincipleJEE AdvancedEasy

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 .........

Answer:9

Visualized Solution (English)

Visualizing the Setup

Given: 4 balls and 4 boxes of matching colors.
Constraint: No ball should go into the box of its own color.
This is a classic problem of Derangements.

Defining Derangement

A Derangement of $n$ objects is a permutation with no fixed points.
Symbolized as $D_{n}$ , it counts the ways to arrange $n$ items so that none are in their original position.
For this problem, we need to find $D_{4}$ .

The Derangement Formula

The general formula for $D_{n}$ is:
$D_{n} = n! (\frac{1}{0 !} - \frac{1}{1 !} + \frac{1}{2 !} - \frac{1}{3 !} + \dots + \frac{( - 1 ) ^{n}}{n !})$
This formula uses the Inclusion-Exclusion Principle.

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n = 4

Substitute $n = 4$ into the formula:
$D_{4} = 4! (\frac{1}{0 !} - \frac{1}{1 !} + \frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !})$

Simplifying the Series

Simplify the factorial terms:
$D_{4} = 4! (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24})$
Notice that $1 - 1 = 0$ , so these terms cancel out.

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4!

Calculate $4!$ :
$4! = 4 \times 3 \times 2 \times 1 = 24$
The expression becomes: $24 (\frac{1}{2} - \frac{1}{6} + \frac{1}{24})$

Final Atomic Compute

Distribute $24$ into the brackets:
$D_{4} = \frac{24}{2} - \frac{24}{6} + \frac{24}{24}$
$D_{4} = 12 - 4 + 1 = 9$

Conclusion and Takeaway

Final Answer: The number of ways is 9.
Key Takeaway: For $n = 4$ , $D_{4} = 9$ .
Pro-tip: Memorizing small derangement values like $D_{3} = 2$ , $D_{4} = 9$ , and $D_{5} = 44$ can save time in JEE.

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