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 T_1 \cap T_2 \cap T_3 \cap T_4 is

Probability PYQ — JEE Advanced 2018 · Shift 1 · Q36 with Solution

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Correct answer: C

Let us count the number of seatings of the five students such that for each consecutive pair

(S_2,S_3), (S_3,S_4), (S_4,S_5),$$ the two students are **not adjacent**. Total number of random seatings of 5 students in 5 seats is $$5! = 120.$$ We need the number of permutations of $S_1,S_2,S_3,S_4,S_5$ in which no two consecutive labels sit next to each other. --- ## 1. Define bad events Let - $A_1$: $S_1$ and $S_2$ are adjacent, - $A_2$: $S_2$ and $S_3$ are adjacent, - $A_3$: $S_3$ and $S_4$ are adjacent, - $A_4$: $S_4$ and $S_5$ are adjacent. We want $$P(T_1\cap T_2\cap T_3\cap T_4)=1-P(A_1\cup A_2\cup A_3\cup A_4).$$ So we use Inclusion-Exclusion. --- ## 2. Count single bad events For any one event, say $A_1$, treat $(S_1,S_2)$ as one block. Inside the block, order can be either $S_1S_2$ or $S_2S_1$. Thus number of arrangements: $$2\cdot 4! = 48.$$ Hence, $$|A_1|=|A_2|=|A_3|=|A_4|=48.$$ So $$\sum |A_i| = 4\cdot 48 = 192.$$ --- ## 3. Count pairwise intersections We must consider two types. ### (i) Adjacent-index events sharing a student Example: $A_1\cap A_2$ means $S_1,S_2,S_3$ must appear as three consecutive students with $S_2$ in the middle: $$S_1S_2S_3 \quad \text{or} \quad S_3S_2S_1.$$ So this triple forms a block with 2 possible internal orders. Then total arrangements: $$2\cdot 3! = 12.$$ Similarly, $$|A_1\cap A_2|=|A_2\cap A_3|=|A_3\cap A_4|=12.$$ Total from these 3 cases: $$3\cdot 12 = 36.$$ ### (ii) Disjoint pairs Example: $A_1\cap A_3$ means $(S_1,S_2)$ adjacent and $(S_3,S_4)$ adjacent. Each pair is a block with 2 internal orders. So we have 3 objects: block 1, block 2, and $S_5$. Thus number of arrangements: $$2\cdot 2\cdot 3! = 24.$$ The disjoint pairs are: $$(A_1,A_3), (A_1,A_4), (A_2,A_4),$$ so total: $$3\cdot 24 = 72.$$ Hence, $$\sum |A_i\cap A_j| = 36+72=108.$$ --- ## 4. Count triple intersections Again, two types. ### (i) Three consecutive events Example: $A_1\cap A_2\cap A_3$ means $S_1,S_2,S_3,S_4$ must be consecutive in either order: $$S_1S_2S_3S_4 \quad \text{or} \quad S_4S_3S_2S_1.$$ So count: $$2\cdot 2! = 4.$$ Similarly for $A_2\cap A_3\cap A_4$: $$4.$$ Total from these two: $$8.$$ ### (ii) One chain of length 2 and one disjoint pair Example: $A_1\cap A_2\cap A_4$. Then $S_1,S_2,S_3$ form a block with orders $$S_1S_2S_3 \text{ or } S_3S_2S_1,$$ and $(S_4,S_5)$ form a block with 2 orders. So total arrangements: $$2\cdot 2\cdot 2! = 8.$$ Similarly, $$|A_1\cap A_2\cap A_4|=8, \qquad |A_1\cap A_3\cap A_4|=8.$$ Total from these two: $$16.$$ Hence, $$\sum |A_i\cap A_j\cap A_k| = 8+16=24.$$ --- ## 5. Count all four intersections For $$A_1\cap A_2\cap A_3\cap A_4,$$ all consecutive pairs must be adjacent, so all five students must sit in one chain: $$S_1S_2S_3S_4S_5 \quad \text{or} \quad S_5S_4S_3S_2S_1.$$ Thus, $$|A_1\cap A_2\cap A_3\cap A_4| = 2.$$ --- ## 6. Apply Inclusion-Exclusion Therefore, $$|A_1\cup A_2\cup A_3\cup A_4| = 192 - 108 + 24 - 2 = 106.$$ So the number of good arrangements is $$120 - 106 = 14.$$ Hence the required probability is $$\frac{14}{120} = \frac{7}{60}.$$ --- ## 7. Check options The probability is $$\boxed{\frac{7}{60}}$$ which corresponds to **Option C**.

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Correct answer: C

Let us count the number of seatings of the five students such that for each consecutive pair

(S_2,S_3), (S_3,S_4), (S_4,S_5),$$ the two students are **not adjacent**. Total number of random seatings of 5 students in 5 seats is $$5! = 120.$$ We need the number of permutations of $S_1,S_2,S_3,S_4,S_5$ in which no two consecutive labels sit next to each other. --- ## 1. Define bad events Let - $A_1$: $S_1$ and $S_2$ are adjacent, - $A_2$: $S_2$ and $S_3$ are adjacent, - $A_3$: $S_3$ and $S_4$ are adjacent, - $A_4$: $S_4$ and $S_5$ are adjacent. We want $$P(T_1\cap T_2\cap T_3\cap T_4)=1-P(A_1\cup A_2\cup A_3\cup A_4).$$ So we use Inclusion-Exclusion. --- ## 2. Count single bad events For any one event, say $A_1$, treat $(S_1,S_2)$ as one block. Inside the block, order can be either $S_1S_2$ or $S_2S_1$. Thus number of arrangements: $$2\cdot 4! = 48.$$ Hence, $$|A_1|=|A_2|=|A_3|=|A_4|=48.$$ So $$\sum |A_i| = 4\cdot 48 = 192.$$ --- ## 3. Count pairwise intersections We must consider two types. ### (i) Adjacent-index events sharing a student Example: $A_1\cap A_2$ means $S_1,S_2,S_3$ must appear as three consecutive students with $S_2$ in the middle: $$S_1S_2S_3 \quad \text{or} \quad S_3S_2S_1.$$ So this triple forms a block with 2 possible internal orders. Then total arrangements: $$2\cdot 3! = 12.$$ Similarly, $$|A_1\cap A_2|=|A_2\cap A_3|=|A_3\cap A_4|=12.$$ Total from these 3 cases: $$3\cdot 12 = 36.$$ ### (ii) Disjoint pairs Example: $A_1\cap A_3$ means $(S_1,S_2)$ adjacent and $(S_3,S_4)$ adjacent. Each pair is a block with 2 internal orders. So we have 3 objects: block 1, block 2, and $S_5$. Thus number of arrangements: $$2\cdot 2\cdot 3! = 24.$$ The disjoint pairs are: $$(A_1,A_3), (A_1,A_4), (A_2,A_4),$$ so total: $$3\cdot 24 = 72.$$ Hence, $$\sum |A_i\cap A_j| = 36+72=108.$$ --- ## 4. Count triple intersections Again, two types. ### (i) Three consecutive events Example: $A_1\cap A_2\cap A_3$ means $S_1,S_2,S_3,S_4$ must be consecutive in either order: $$S_1S_2S_3S_4 \quad \text{or} \quad S_4S_3S_2S_1.$$ So count: $$2\cdot 2! = 4.$$ Similarly for $A_2\cap A_3\cap A_4$: $$4.$$ Total from these two: $$8.$$ ### (ii) One chain of length 2 and one disjoint pair Example: $A_1\cap A_2\cap A_4$. Then $S_1,S_2,S_3$ form a block with orders $$S_1S_2S_3 \text{ or } S_3S_2S_1,$$ and $(S_4,S_5)$ form a block with 2 orders. So total arrangements: $$2\cdot 2\cdot 2! = 8.$$ Similarly, $$|A_1\cap A_2\cap A_4|=8, \qquad |A_1\cap A_3\cap A_4|=8.$$ Total from these two: $$16.$$ Hence, $$\sum |A_i\cap A_j\cap A_k| = 8+16=24.$$ --- ## 5. Count all four intersections For $$A_1\cap A_2\cap A_3\cap A_4,$$ all consecutive pairs must be adjacent, so all five students must sit in one chain: $$S_1S_2S_3S_4S_5 \quad \text{or} \quad S_5S_4S_3S_2S_1.$$ Thus, $$|A_1\cap A_2\cap A_3\cap A_4| = 2.$$ --- ## 6. Apply Inclusion-Exclusion Therefore, $$|A_1\cup A_2\cup A_3\cup A_4| = 192 - 108 + 24 - 2 = 106.$$ So the number of good arrangements is $$120 - 106 = 14.$$ Hence the required probability is $$\frac{14}{120} = \frac{7}{60}.$$ --- ## 7. Check options The probability is $$\boxed{\frac{7}{60}}$$ which corresponds to **Option C**.

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