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(C) The total number of ways to place $6$ letters in $6$ envelopes is $6!$.Let $S$ be the total number of ways,which is $6!$.Let $E_0$ be the case whe

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$\sum_{r=2}^6{ }^6 C_{6-r} \cdot D_r$

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(C) The total number of ways to place $6$ letters in $6$ envelopes is $6!$.Let $S$ be the total number of ways,which is $6!$.Let $E_0$ be the case where all letters are in the correct envelopes ($1$ way).Let $E_1$ be the case where exactly one letter is in the wrong envelope. This is impossible,as if $5$ are correct,the $6^{th}$ must also be correct.Therefore,the number of ways where at least two letters are in the wrong envelopes is the total ways minus the ways where $0$ or $1$ letter is in the wrong envelope.This is equivalent to $6! - (1 + 0) = 6! - 1$.Alternatively,using the derangement formula $D_r$ (where $r$ is the number of letters in wrong envelopes),we sum the cases where $r$ letters are in wrong envelopes for $r = 2, 3, 4, 5, 6$.The number of ways is $\sum_{r=2}^6 { }^6 C_{6-r} D_r = { }^6 C_4 D_2 + { }^6 C_3 D_3 + { }^6 C_2 D_4 + { }^6 C_1 D_5 + { }^6 C_0 D_6$.

$A$ person writes letters to $6$ friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?
Notation : $D_n = n! \left( \sum_{i=0}^n \frac{(-1)^i}{i!} \right)$

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$A$ person writes letters to $6$ friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?Notation : $D_n = n! \left( \sum_{i=0}^n \frac{(-1)^i}{i!} \right)$