1.

There are five students S_(1),S_(2),S_(3),S_(4) and R_(5) arranged in a row, where initially the seat R_(1)is allotted to the students are randomly allotted the five seats R_(1),R_(2),R_(3),R_(4) and R_(6) arranged in a row, where initially the seat R_(i) is allotted to the student S_(i)i,=1,2,3,4,5. But, on the examination day, the five students are randomly allotted the five seats. (There are two questions based on Paragraph, the question given below is one of them) The probability that, on the examination day, the student S_(1) gets the previously allotted seat R_(1), and NONE of the remaining students gets the seat previously allotted to him/her is

Answer»

`(3)/(40)`
`(1)/(8)`
`(7)/(40)`
`(1)/(5)`

Solution :Here five students `S_(1),S_(2),S_(3),S_(4) and S_(5)` and five seats `R_(1),R_(2),R_(3),R_(4) and R_(5)`
`:.` TOTAL number of arrangement of sitting five students is `5!=120`.
Here, `S_(1)` gets previouslyalloted SEAT `R_(1)` .
`:. S_(2) ,S_(3),S_(4) and S_(5)` not get previously seats .
Totalnumber of WAY `S_(2) , S_(3), S_(4) and S_(5)` not get previously seat is
`4! (1- (1)/(1!) +(1)/(2!) -(1)/(3!) +(1)/(4!))=24(1-1+(1)/(2)-(1)/(6)+(1)/(24))=24((12-4+1)/(24))=9`
`:.` Required probability `=(9)/(!20)=(3)/(40)`


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