Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma1630: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Ma1630 Solution: There are five students S 1 , S 2 , S 3 , S 4 and S 5 in a…

There are five students S1, S2, S3, S4 and S5 in a music class and for them there are five seats R1, R2, R3, R4 and R5 arranged in a row, where initially the seat Ri is allotted to the student Si, 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 "A", the question given below is one of them)

The probability that, on the examination day, the student S1 gets the previously allotted seat R1, and NONE of the remaining students gets the seat previously allotted to him/her is

Multiple Choice Options

  • A 340{3 \over {40}}
  • B 18{1 \over 8}
  • C 740{7 \over 40}
  • D 15{1 \over 5}

Step-by-Step Verified Solution

Here, five students S1, S1, S3, S4 and S5 and five seats R1, R2, R3, R4 and R5

∴\therefore Total number of arrangement of sitting five students is 5! = 120

Here, S1 gets previously allotted seat R1

∴\therefore S2, S3, S4 and S5 not get previously seats.

Total number of way S2, S3, S4 and S5 not get previously seats is

4!(1−11!+12!−13!+14!)4!\left( {1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}} \right)

=24(1−1+12−16+124)= 24\left( {1 - 1 + {1 \over 2} - {1 \over 6} + {1 \over {24}}} \right)

=24(12−4+124)=9= 24\left( {{{12 - 4 + 1} \over {24}}} \right) = 9

∴\therefore Required probability = 9120=340{9 \over {120}} = {3 \over {40}}

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