Permutations and Combinations — JEE Advanced · 2018 · 20 May 2018 · Paper 1 Question with Solution

Passage

There are five students

S_{1}, S_{2}, S_{3}, S_{4}

and

S_{5}

in a music class and for them there are five seats

R_{1}, R_{2}, R_{3}, R_{4}

and

R_{5}

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 five students

S_{1}, S_{2}, S_{3}, S_{4}

and

S_{5}

in a music class and for them there are five seats

R_{1}, R_{2}, R_{3}, R_{4}

and

R_{5}

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.

For

i = 1, 2, 3, 4

, let

T_{i}

denote the event that the students

S_{i}

and

S_{i + 1}

do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event

T_{1} \cap T_{2} \cap T_{3} \cap T_{4}

Answer: (C)

Solution

Total number of seatings:

5! = 120

.

Let

A_{i}

(

i = 1, 2, 3, 4

) be the event that

S_{i}

and

S_{i + 1}

do sit together. We want the seatings in none of the

A_{i}

; by inclusion–exclusion,

∣ A_{1} \cup A_{2} \cup A_{3} \cup A_{4} ∣ = \sum ∣ A_{i} ∣ - \sum ∣ A_{i} \cap A_{j} ∣ + \sum ∣ A_{i} \cap A_{j} \cap A_{k} ∣ - ∣ A_{1} \cap A_{2} \cap A_{3} \cap A_{4} ∣.

$∣ A_{i} ∣ = 2 \cdot 4! = 48$ (tie the pair into one block, 2 internal orders), so $\sum ∣ A_{i} ∣ = 4 \times 48 = 192$ .
Two events: if the pairs share a student ( $A_{1} \cap A_{2}$ , $A_{2} \cap A_{3}$ , $A_{3} \cap A_{4}$ ) they form a block of three with 2 internal orders, giving $2 \cdot 3! = 12$ each; if disjoint ( $A_{1} \cap A_{3}$ , $A_{1} \cap A_{4}$ , $A_{2} \cap A_{4}$ ) two blocks of two give $2 \cdot 2 \cdot 3! = 24$ each. $\sum ∣ A_{i} \cap A_{j} ∣ = 3 \times 12 + 3 \times 24 = 108$ .
Three events: $A_{1} \cap A_{2} \cap A_{3}$ and $A_{2} \cap A_{3} \cap A_{4}$ make a block of four ( $2 \cdot 2! = 4$ each); $A_{1} \cap A_{2} \cap A_{4}$ and $A_{1} \cap A_{3} \cap A_{4}$ make a block of three and a block of two ( $2 \cdot 2 \cdot 2! = 8$ each). $\sum = 4 + 4 + 8 + 8 = 24$ .
All four: one block of five, $2$ orders.

∣ A_{1} \cup A_{2} \cup A_{3} \cup A_{4} ∣ = 192 - 108 + 24 - 2 = 106

, and the favourable seatings number

120 - 106 = 14

.

Listed out (seat

R_{1}

R_{5}

left to right), the 14 seatings are:

S_{1} S_{3} S_{5} S_{2} S_{4}

S_{1} S_{4} S_{2} S_{5} S_{3}

S_{2} S_{4} S_{1} S_{3} S_{5}

S_{2} S_{4} S_{1} S_{5} S_{3}

S_{2} S_{5} S_{3} S_{1} S_{4}

S_{3} S_{1} S_{4} S_{2} S_{5}

S_{3} S_{1} S_{5} S_{2} S_{4}

S_{3} S_{5} S_{1} S_{4} S_{2}

S_{3} S_{5} S_{2} S_{4} S_{1}

S_{4} S_{1} S_{3} S_{5} S_{2}

S_{4} S_{2} S_{5} S_{1} S_{3}

S_{4} S_{2} S_{5} S_{3} S_{1}

S_{5} S_{2} S_{4} S_{1} S_{3}

S_{5} S_{3} S_{1} S_{4} S_{2}

P (T_{1} \cap T_{2} \cap T_{3} \cap T_{4}) = \frac{14}{120} = \frac{7}{60} .

Answer: (C)

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Single correct option