Six cards and six envelopes are numbered 1, 2 | vedclass

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(C) Let the cards be $C_1, C_2, C_3, C_4, C_5, C_6$ and envelopes be $E_1, E_2, E_3, E_4, E_5, E_6$. Given $C_1$ is in $E_2$. We need to place $C_2, C

Vedclass · Accepted Answer

$53$

Vedclass · Answer

(C) Let the cards be $C_1, C_2, C_3, C_4, C_5, C_6$ and envelopes be $E_1, E_2, E_3, E_4, E_5, E_6$. Given $C_1$ is in $E_2$. We need to place $C_2, C_3, C_4, C_5, C_6$ in $E_1, E_3, E_4, E_5, E_6$ such that $C_i$ is not in $E_i$ for $i \in \{2, 3, 4, 5, 6\}$. Case $1$: $C_2$ is in $E_1$. Then we need to place $C_3, C_4, C_5, C_6$ in $E_3, E_4, E_5, E_6$ such that no card $C_i$ is in $E_i$. This is a derangement of $4$ objects,$D_4 = 4!(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}) = 24(\frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 12 - 4 + 1 = 9$. Case $2$: $C_2$ is not in $E_1$. We have $5$ cards $C_2, C_3, C_4, C_5, C_6$ to be placed in $E_1, E_3, E_4, E_5, E_6$ such that $C_2 
eq E_1, C_3 
eq E_3, C_4 
eq E_4, C_5 
eq E_5, C_6 
eq E_6$. This is a derangement of $5$ objects,$D_5 = 5!(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}) = 120(\frac{60-20+5-1}{120}) = 44$. Total ways = $9 + 44 = 53$.

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