Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so

Q: Wait, why can we just divide by 5? Isn't that too simple for a JEE problem?

It feels like a shortcut, but it is rooted in the fundamental symmetry of the derangement set. Because the condition 'no card in its own envelope' applies to all cards equally, the set of all derangements is perfectly symmetric with respect to the position of any specific card. Dividing by $(n-1)$ is mathematically rigorous here because Card 1 has exactly $(n-1)$ valid positions it can occupy in any derangement, and each position is equally likely.

Q: Can I solve this using the Principle of Inclusion-Exclusion (PIE) with the constraint $C_1 \in E_2$?

Absolutely! You can fix $C_1$ in $E_2$ and then count the remaining derangements. However, it involves splitting into two cases: $C_2$ in $E_1$ and $C_2$ not in $E_1$. It is much more tedious and prone to calculation errors. Symmetry is your best friend in competitive exams, and recognizing it saves you precious time.

Q: What if the question asked for the number of ways where Card 1 is in Envelope 1?

That would be zero! By the very definition of a derangement, no card can ever be in its own envelope. If you ever find yourself calculating a case where $C_i \in E_i$, you know you've hit a contradiction. Always keep the fundamental constraint of the problem at the forefront of your mind.

Question

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is

Options:

(A) 264

(B) 265

(C) 53

(D) 67

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to find the number of ways to arrange six cards into six envelopes such that no card is in its own envelope (a derangement), with the additional specific constraint that Card 1 must be placed in Envelope 2.

This is a classic problem of **Derangements**, denoted as $D_n$, where $n$ items are permuted such that no item occupies its original position.

### The Master Equation for Derangements

Before addressing the specific constraint, we must calculate the total number of derangements for $n=6$. The formula, derived from the **Principle of Inclusion-Exclusion**, is:

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

For $n=6$, we substitute the values into the formula:

$$D_6 = 720 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + \frac{1}{720} ight)$$

Simplifying the expression by noting that $1 - 1 = 0$, we obtain:

$$D_6 = 720 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + \frac{1}{720} ight)$$

Distributing the $720$ across the terms yields:

$$D_6 = 360 - 120 + 30 - 6 + 1 = 265$$

Thus, there are exactly $265$ total ways to arrange the six cards such that no card is in its matching envelope.

### The Symmetry Insight

Now, we introduce the constraint: Card 1 must be in Envelope 2. In any valid derangement, Card 1 cannot be in Envelope 1.

Therefore, Card 1 must occupy one of the remaining five envelopes: $E_2, E_3, E_4, E_5,$ or $E_6$. Because the condition "no card in its own envelope" applies to all cards equally, there is no mathematical distinction between Card 1 being in any of these five envelopes.

The constraints are perfectly symmetric. Consequently, the total number of derangements $D_6$ is distributed equally among these $5$ possible positions for Card 1. If $N$ is the number of ways Card 1 is in Envelope 2, then:

$$D_6 = 5 	imes N$$

### Final Calculation

Using our total $D_6 = 265$ and the symmetry relation $N = \frac{D_6}{5}$, we perform the final arithmetic:

$$N = \frac{265}{5} = 53$$

The total number of ways to satisfy the given conditions is **53**. This result demonstrates that recognizing symmetry is often more powerful than brute-force calculation in competitive mathematics.

Select Answer:

Visualized Solution

$6$ Cards and $6$ Envelopes

Constraint 1: Derangement

The Derangement Formula $D_{n}$

Applying Formula for $n = 6$

Expanding the Terms

Calculating Total Derangements $D_{6}$

Constraint 2: Fixing $C_{1}$

The Symmetry Principle

Applying Symmetry

Final Answer

The Sigma Insight: Derangement Principle

Analyzing the Setup

The Master Equation for Derangements

The Symmetry Insight

Final Calculation

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