ITPEC FE Subject A October 2025 Question 3
Three friends and 4 other people are randomly seated in 7 seats arranged in a row. What is the probability that the 3 friends sit next to one another?
Explanation
1/7 — treat the three friends as one block, count the seatings that keep the block together, and divide by all seatings.
- •All seatings of 7 people in 7 seats: 7! = 5,040
- •Positions for 3 adjacent seats in a row of 7: seats 1–3 through 5–7, so 5
- •Orders of the friends inside the block: 3! = 6
- •Orders of the other 4 people in the remaining seats: 4! = 24
- •Favourable seatings: 5 × 6 × 24 = 720
- •Probability: 720 ÷ 5,040 = 1/7
A shorter check: the friends' seats are one of C(7, 3) = 35 equally likely sets of 3 seats, and 5 of those sets are adjacent, so 5/35 = 1/7.
Answer (d)
Why not others:
- •(c) 1/35 — counts only one block of seats out of the 35 seat sets, forgetting that the block can start in 5 places
- •(b) 1/42 — 5 × 4! = 120 and 120 ÷ 5,040: forgets the 3! orders of the friends inside the block
- •(a) 1/168 — 5 × 3! = 30 and 30 ÷ 5,040: forgets the 4! orders of the other people
Key rule: For "these k people stay together", glue them into one block: block positions × orders inside the block × orders of everyone else, over the unrestricted total.
Basic Theory · AI-written, may contain errors
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