Five persons P, Q, R, S and T are sitting in a row, not necessarily in the same order. Q and R are separated by one person, and S should not be seated adjacent to Q. The number of distinct seating arrangements possible is: A. 8 B. 8 C. 10 D. 16

GATE 2021 · General Aptitude · Seating Arrangement · medium

Answer: 16 distinct seating arrangements (Option D)

1. List all (Q, R) placements separated by exactly one person: Valid ordered pairs (Q position, R position): (1,3), (2,4), (3,5), (3,1), (4,2), (5,3). That is 6 cases. For each, the gap seat is the middle of Q and R.
2. Count valid completions for each of the 6 cases: Case (Q=1,R=3): gap at 2; remaining seats {2,4,5} for {P,S,T}; Q adjacent to seat 2 only (seat 0 doesn't exist). S cannot be at seat 2: 3! - 2! = 6 - 2 = 4. Case (Q=2,R=4): gap at 3; remaining seats {1,3,5} for {P,S,T}; Q adjacent to seats 1 and 3. S can only go to seat 5; P,T fill {1,3} in 2! = 2 ways. Valid = 2. Case (Q=3,R=5): gap at 4; remaining seats {1,2,4} for {P,S,T}; Q adjacent to seats 2 and 4. S can only go to seat 1; P,T fill {2,4} in 2 ways. Valid = 2. Case (Q=3,R=1): gap at 2; remaining seats {2,4,5} for {P,S,T}; Q adjacent to seats 2 and 4. S cannot be at 2 or 4, so S must be at 5; P,T fill {2,4} in 2 ways. Valid = 2. Case (Q=4,R=2): gap at 3; remaining seats {1,3,5} for {P,S,T}; Q adjacent to seats 3 and 5. S cannot be at 3 or 5, so S must be at 1; P,T fill {3,5} in 2 ways. Valid = 2. Case (Q=5,R=3): gap at 4; remaining seats {1,2,4} for {P,S,T}; Q adjacent to seat 4 only (seat 6 doesn't exist). S cannot be at 4; S can be at 1 or 2 (2 choices); for each, P and T fill the remaining 2 seats in 2! = 2 ways. Valid = 2 * 2 = 4.
3. Sum all valid counts: Total = 4 + 2 + 2 + 2 + 2 + 4 = 16.