There are 6 rooms on the first floor of a hotel, with 3 rooms on each side of the corridor, each room exactly opposite to the other room. The number of ways in which three guests are accommodated in three of the six rooms, one in each room, such that no two guests are in adjacent rooms or in the opposite rooms is

Step-by-Step Solution:

Step 1: Define the problem

We need to select 3 rooms out of the 6 such that the selected rooms satisfy the conditions. Adjacent rooms are defined as rooms next to each other on the same side (e.g., Room A and Room B), and opposite rooms are defined as rooms directly facing each other across the corridor (e.g., Room A and Room D).

Step 2: Eliminate invalid combinations

We begin by eliminating combinations where any two rooms are adjacent or opposite:

• Adjacent rooms on Side 1: (A, B), (B, C)
• Adjacent rooms on Side 2: (D, E), (E, F)
• Opposite rooms: (A, D), (B, E), (C, F)

Step 3: Count valid combinations

Next, we select 3 rooms such that no two rooms are adjacent or opposite. The valid combinations can be identified as follows:

• Select one room from Side 1 and two rooms from Side 2, or vice versa.
• The selected rooms must not be adjacent or opposite.

Case 1: Select one room from Side 1 and two rooms from Side 2

• From Side 1, valid choices: A or C (since B is adjacent to both A and C).
• From Side 2, valid choices depend on the room selected from Side 1:
  • If A is selected, valid rooms from Side 2 are E and F (D is opposite to A).
  • If C is selected, valid rooms from Side 2 are D and E (F is opposite to C).

Total combinations for Case 1:

• If A is selected: {A, E, F}
• If C is selected: {C, D, E}

Number of combinations = 2.

Case 2: Select two rooms from Side 1 and one room from Side 2

• From Side 1, valid choices depend on avoiding adjacent rooms:
  • Valid pairs: {A, C} (B is adjacent to both A and C).
• From Side 2, the room must not be opposite to any selected room from Side 1:
  • If {A, C} is selected, valid room from Side 2 is E (D is opposite to A, and F is opposite to C).

Total combinations for Case 2:

• {A, C, E}

Number of combinations = 1.

Total Number of Valid Combinations:

Adding the combinations from both cases:

Case 1: 2 combinations

Case 2: 1 combination

Total = 2 + 1 = 3 combinations.

Step 4: Factorial arrangements within rooms

Since each guest can be assigned to any of the selected rooms, we calculate permutations for arranging 3 guests in 3 rooms:

Permutations = 3 × 2 × 1 = 6.

Final Calculation:

Total arrangements = Number of valid combinations × Permutations

Total arrangements = 3 × 6 = 18.

Correct Answer: Option 4 (18)