Question

How many managers would get their rooms as per their best preference?

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5
• A. 5
• B. 6
• C. 7
• D. 8
• J. 9
• A. 5
• B. 6
• C. 7
• D. 8
• O. 9

Answer 1

The Correct Option is C

Step 1: Recall the preference table. \[ \begin{array}{|c|c|c|c|c|c|} \hline

Preference &

M1 &

M2 &

M3 &

M4 &

M5
\hline 1^\text{st} & 302 & 302 & 303 & 302 & 301
2^\text{nd} & 303 & 304 & 301 & 305 & 302
3^\text{rd} & 304 & 305 & 304 & 304 & 305
4^\text{th} & 301 & 305 & 305 & 303 &
5^\text{th} & & & 302 & &
\hline \end{array} \]

Step 2: Identify unique allocations by first preference.
- Room 301: preferred only by M5. Hence, M5 surely gets Room 301. ✓
- Room 303: preferred only by M3. Hence, M3 surely gets Room 303. ✓
- Room 302: preferred by M1, M2, M4 (conflict; only one can get).


Step 3: Assign conflict-free rooms.
- M5 gets Room 301 (best choice). ✓
- M3 gets Room 303 (best choice). ✓


Step 4: Handle Room 302 conflict.
Among M1, M2, and M4, only one can be allotted Room 302 as per best choice. ✓

Step 5: Count managers satisfied with best choice.
Thus, M3, M5, and one among \{M1, M2, M4\} get their best choice = total 3. \[ \boxed{3} \]

Answer 2

Step 1: Start with certain allocations.
- M5 gets Room 301 (1st preference, rank = 1).
- M3 gets Room 303 (1st preference, rank = 1).

Step 2: Deal with Room 302 conflict (M1, M2, M4).
Only one of them can get Room 302 as 1st preference. Suppose M1 gets it (rank = 1).
Now, M2 and M4 cannot get 302; they move to next available preferences.

Step 3: Assign second preferences.
- M2’s 2nd choice is 304 → assign Room 304 (rank = 2).
- M4’s 2nd choice is 305 → assign Room 305 (rank = 2).

Step 4: Verify uniqueness of allocations.
Now each room 301–305 is allocated uniquely:
M1 → 302, M2 → 304, M3 → 303, M4 → 305, M5 → 301.

Step 5: Compute total ranking sum.
M1 (1) + M2 (2) + M3 (1) + M4 (2) + M5 (1) = 7. \[ \boxed{7} \]

Answer 3

Step 1: Write down the new preference table (M1, M3, M4, M5, M6). \[ \begin{array}{|c|c|c|c|c|c|} \hline

Preference &

M1 &

M3 &

M4 &

M5 &

M6
\hline 1^\text{st} & 302 & 303 & 302 & 301 & 301
2^\text{nd} & 303 & 301 & 305 & 302 & 302
3^\text{rd} & 304 & 304 & 304 & 305 & 303
4^\text{th} & 301 & 305 & 303 & & 304
5^\text{th} & 302 & & & &
\hline \end{array} \]

Step 2: Handle clear allocations.
- M3’s 1st preference is Room 303 → no conflict (allocate 303). Rank = 1. ✓


Step 3: Deal with conflicts.
- Room 301 is preferred 1st by M5 and M6. Only one of them can get 301.
- Room 302 is preferred 1st by M1 and M4 (and also 2nd by M6). Only one of M1/M4 can get it.

Step 4: Assign optimal allocation to minimize total.
- Assign M6 → Room 301 (1st preference). Rank = 1. ✓
- Assign M5 → Room 305 (3rd preference, since 301 and 302 are taken). Rank = 3. ✓
- Assign M1 → Room 302 (1st preference). Rank = 1. ✓
- Assign M4 → Room 304 (3rd preference, since 302 taken). Rank = 3. ✓


Step 5: Add up preference rankings.
M1 (1) + M3 (1) + M4 (3) + M5 (3) + M6 (1) = 9. \[ \boxed{9} \]