Derangement Problem of Cards in Numbered Envelopes

To determine the number of ways the cards can be placed into the envelopes, we define the placement of cards into envelopes as a permutation $f$ of the set $\{1, 2, 3, 4, 5, 6\}$, where $f(i)$ represents the envelope into which card $i$ is placed.

The problem imposes the following conditions:

1. Each envelope contains precisely one card, so $f$ is a bijection (a permutation of $6$ elements).
2. No card is placed into the envelope bearing its own number, which means $f(i) \neq i$ for all $i \in \{1, 2, 3, 4, 5, 6\}$. This implies that $f$ is a derangement of $6$ elements.
3. Card $1$ is specifically placed into envelope $2$, which means $f(1) = 2$.

The total number of derangements of $n$ elements, denoted by $D_n$, is given by the formula: $D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$

For $n = 6$: $D_6 = 6! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} \right)$ $D_6 = 720 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} + \frac{1}{720} \right) = 360 - 120 + 30 - 6 + 1 = 265$

In any derangement of $6$ elements, the first card, card $1$, can be placed into any envelope other than envelope $1$. Thus, $f(1) \in \{2, 3, 4, 5, 6\}$.

By the inherent symmetry of derangements, the number of derangements in which $f(1) = k$ is identical for each $k \in \{2, 3, 4, 5, 6\}$.

Alternatively, using the standard recurrence relation for derangements: $D_n = (n - 1)(D_{n-1} + D_{n-2})$ When card $1$ is fixed into envelope $2$, there are two mutually exclusive cases for card $2$:

• Case 1: Card $2$ goes into envelope $1$ ($f(2) = 1$). The remaining $4$ cards must be deranged among the remaining $4$ envelopes, which can be done in: $D_4 = 4! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 12 - 4 + 1 = 9 \text{ ways}$

• Case 2: Card $2$ does not go into envelope $1$ ($f(2) \neq 1$). Card $2$ cannot go into envelope $1$, card $3$ cannot go into envelope $3$, and so forth. This is equivalent to finding the number of derangements of $5$ objects, which can be done in: $D_5 = 5! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) = 60 - 20 + 5 - 1 = 44 \text{ ways}$

Therefore, the total number of ways to place card $1$ into envelope $2$ while satisfying the derangement condition is: $\text{Total ways} = D_4 + D_5 = 9 + 44 = 53$

Equivalently, using symmetry: $\text{Total ways} = \frac{D_6}{5} = \frac{265}{5} = 53$

Hence, the correct option is C.