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(B) Let $S = \{1, 2, 3, 4, 5\}$. We want to find the number of permutations $\sigma$ of $S$ such that at least $3$ elements are not in their original

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$109$

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(B) Let $S = \{1, 2, 3, 4, 5\}$. We want to find the number of permutations $\sigma$ of $S$ such that at least $3$ elements are not in their original positions.This is equivalent to saying that at most $2$ elements are in their original positions.Let $A_i$ be the set of permutations where the $i$-th digit is in its original position.We want to find the number of permutations where at most $2$ digits are fixed.Total permutations = $5! = 120$.Let $f(k)$ be the number of permutations with exactly $k$ fixed points. $f(k) = \binom{5}{k} D_{5-k}$,where $D_n$ is the number of derangements of $n$ objects.$D_0 = 1, D_1 = 0, D_2 = 1, D_3 = 2, D_4 = 9, D_5 = 44$.The number of permutations with at least $3$ digits not in their original positions is the total permutations minus those with $4$ or $5$ fixed points.Permutations with $5$ fixed points: $\binom{5}{5} D_0 = 1 	imes 1 = 1$.Permutations with $4$ fixed points: $\binom{5}{4} D_1 = 5 	imes 0 = 0$.Permutations with $3$ fixed points: $\binom{5}{3} D_2 = 10 	imes 1 = 10$.Number of permutations with at most $2$ fixed points = $120 - (1 + 0 + 10) = 120 - 11 = 109$.

The number of arrangements of all digits of $12345$ such that at least $3$ digits will not come in their original positions is

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