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

$\text{Given: cards } 1,2,3,4,5,6 \ \to \ \text{envelopes } 1,2,3,4,5,6 \ \text{(one each)}$ $\text{no card } i \text{ in envelope } i$ $1 \to \text{envelope } 2 \quad (\text{fixed})$ $\text{To find: number of such arrangements}$ $1.\ \text{fix card } 1 \to \text{envelope } 2$ $2.\ \text{arrange the remaining } 5 \text{ cards avoiding their own envelopes}$ $3.\ \text{count with inclusion-exclusion}$ $\text{cards left: } \{2,3,4,5,6\}$ $\text{envelopes left: } \{1,3,4,5,6\}$ $\text{without restriction: } 5! = 120$ $\text{forbidden: card } i \to \text{envelope } i, \quad i = 3,4,5,6$ $4 \text{ forbidden events}$ $N = 5! - \binom{4}{1}4! + \binom{4}{2}3! - \binom{4}{3}2! + \binom{4}{4}1!$ $5! = 120$ $\binom{4}{1}4! = 4 \times 24 = 96$ $\binom{4}{2}3! = 6 \times 6 = 36$ $\binom{4}{3}2! = 4 \times 2 = 8$ $\binom{4}{4}1! = 1$ $N = 120 - 96 + 36 - 8 + 1$ $= 24 + 36 - 8 + 1 = 53$ $!6 = 265, \quad \dfrac{265}{5} = 53 \ \checkmark$ $\boxed{53} \quad \Rightarrow \quad \text{option (3)}$