InterviewSolution
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InterviewSolution
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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 toA. `(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))` |
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Answer» Correct Answer - 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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