Question

Five friends (A, B, C, D, E) sit in a row. A is to the left of B, and C is not at an end. How many arrangements are possible?

• A. 24
  
• B. 36
  
• C. 48
  
• D. 60
  

Hint:

For seating arrangements, calculate total permutations and adjust for each restriction systematically.

Answer 1

The Correct Option is B

- Step 1: Total arrangements without restrictions. 5 people in a row: $5! = 120$.
- Step 2: Apply restriction A to the left of B. In any arrangement, A is left of B in half the cases (symmetry): $\frac{120}{2} = 60$.
- Step 3: Apply restriction C not at ends. Ends are positions 1 and 5 (2 positions). C can be in positions 2, 3, or 4 (3 out of 5 positions). Fraction of arrangements where C is not at ends = $\frac{3}{5}$.
- Step 4: Calculate valid arrangements. Total with A left of B = 60. With C not at ends: $60 \times \frac{3}{5} = 36$.
- Step 5: Alternative approach. Place C in one of 3 middle positions (2, 3, 4): 3 choices. Arrange remaining 4 (A, B, D, E) with A left of B: 4 positions, choose 2 for A and B (order AB), $\binom{4}{2} = 6$, then arrange D, E in 2 positions: $2! = 2$. Total = $3 \times 6 \times 2 = 36$.
- Step 6: Verify. Both methods yield 36. Check options: (1) 24, (2) 36, (3) 48, (4) 60.
- Step 7: Conclusion. Option (2) is correct.