The number of arrangements of all digits of 1 | vedclass

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Question

(B) The total number of arrangements of $5$ distinct digits is $5! = 120$.We need to find the number of arrangements where at least $3$ digits do not

Vedclass · Accepted Answer

$109$

Vedclass · Answer

(B) The total number of arrangements of $5$ distinct digits is $5! = 120$.We need to find the number of arrangements where at least $3$ digits do not appear in their original positions.Let $S$ be the set of all permutations. We use the principle of derangements $(D_n)$.The number of ways to choose $k$ digits that are $NOT$ in their original positions is given by $\binom{5}{k} 	imes D_k$,where $D_k$ is the number of derangements of $k$ objects.For $k=3$: $\binom{5}{3} 	imes D_3 = 10 	imes 2 = 20$.For $k=4$: $\binom{5}{4} 	imes D_4 = 5 	imes 9 = 45$.For $k=5$: $\binom{5}{5} 	imes D_5 = 1 	imes 44 = 44$.Total arrangements = $20 + 45 + 44 = 109$.

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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