Mathematics - Permutation & Combination Question with Solution | TestHub

MathematicsPermutation & CombinationDearrangementEasy2 minQB

MathematicsEasysingle choice

Passage / Comprehension

Let $θ = (a_{1}, a_{2}, a_{3}, \dots, a_{n})$ be a given arrangement of $n$ distinct objects $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ . A derangement of $θ$ is an arrangement of these $n$ objects in which none of the objects occupies its original position. Let $D_{n}$ be the number of derangements of the permutation $θ$ .

$D_{n}$ is equal to

Options:

Answer:

Solution:

For every choice of $r = 1, 2, \dots, (n - 1),$ , when the $n$ ᵗʰ object $a_{n}$ goes to the $r$ ᵗʰ place, there are $D_{n - 1} + D_{n - 2}$ ways for the other $n - 1$ objects $a_{1}, a_{2}, \dots, a_{n - 1}$ to be deranged.

Hence,

$D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})$

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:Dearrangement

⏱ 2mℹ️ Source: QB

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