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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

Answer»

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

Solution :`(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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