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Let theta=(a(1),a(2),a(3),...,a(n)) be a...

Let `theta=(a_(1),a_(2),a_(3),...,a_(n))` be a given arrangement of `n` distinct objects `a_(1),a_(2),a_(3),…,a_(n)`. A derangement of `theta` is an arrangment of these `n` objects in which none of the objects occupies its original position. Let `D_(n)` be the number of derangements of the permutations `theta`.
`D_(n)` is equal to

A

`(n-1)D_(n-1)+D_(n-2)`

B

`D_(n-1)+(n-1)D_(n-2)`

C

`n(D_(n-1)+D_(n-2))`

D

`(n-1)(D_(n-1)+D_(n-2))`

Text Solution

Verified by Experts

The correct Answer is:
D
`(d)` For every choice of `r=1,2,3,….(n-1)` when the `n^(th)` object `a_(n)` goes to the `rth` place, there are `D_(n-1)+D_(n-2)` ways of the other `(n-1)` objects `a_(1)`, `a_(2)`, ….,`a_(n-1)` to be deranged.
Hence `=D_(n)=(n-1)(D_(n-1)+D_(n-2))`
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