Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma424f: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Ma424f 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)

For i = 1, 2, 3, 4, let Ti denote the event that the students Si and Si+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1∩T2∩T3∩T4{T_1} \cap {T_2} \cap {T_3} \cap {T_4} is

Multiple Choice Options

  • A 115{1 \over {15}}
  • B 110{1 \over {10}}
  • C 760{7 \over {60}}
  • D 15{1 \over {5}}

Step-by-Step Verified Solution

Here, n(T1∩T2∩T3∩T4)n({T_1} \cap {T_2} \cap {T_3} \cap {T_4}) = Total −n(T1‾∪T2‾∪T3‾∪T4‾)- n(\overline {{T_1}} \cup \overline {{T_2}} \cup \overline {{T_3}} \cup \overline {{T_4}} )

⇒\Rightarrow n(T1∩T2∩T3∩T4)n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})

=5!−[4C14!2! −(3C13!2!+3C13!2!2!)+(2C12!2! + 4C12.2!)−2]= 5! - [{}^4{C_1}4!2!\, - ({}^3{C_1}3!2! + {}^3{C_1}3!2!2!) + ({}^2{C_1}2!2!\, + \,{}^4{C_1}2.2!) - 2]

⇒\Rightarrow n(T1∩T2∩T3∩T4)n({T_1} \cap {T_2} \cap {T_3} \cap {T_4})

=120−[192−(36+72)+(8+16)−2]= 120 - [192 - (36 + 72) + (8 + 16) - 2]

=120−[192−108+24−2]=14= 120 - [192 - 108 + 24 - 2] = 14

∴\therefore Required probability = 14120=760{{14} \over {120}} = {7 \over {60}}

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