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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`.
There are `5` different colour balls and `5` boxes of colours same as those of the balls. The number of ways in which one can place the balls into the boxes, one each in a box, so that no ball goes to a box of its own colour is

A

`40`

B

`44`

C

`45`

D

`60`

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` `(D_(n))/(n!)-(D_(n-1))/((n-1)!)=((-1)^(n))/(n!)` gives
`(D_(n))/(n!)=sum_(r=2)^(n)((D_(r ))/(r!)-(D_(r-1))/((r-1)!))=sum_(r=2)^(n)((-1)^(r ))/(r!)`
`impliesD_(n)=n!sum_(r=2)^(n)((-1)^(r ))/(r!)`
`:. D_(5)=5!((1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!))=44`
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