Discussion
11 students plan to have dinner together for multiple days. They will be seated at the round table. Their plan is to ensure that each student sits next to different neighbours at every dinner. How many days can this arrangement be sustained ?
Answer: A
The arrangement can be sustained for 5 days.
Model the students as vertices of the complete graph \(K_{11}\); each seating at a round table is a Hamiltonian cycle, and different neighbours each day means the cycles share no edge. Each student has 10 possible neighbours and gains 2 per dinner, so at most 10/2 = 5 dinners are possible.
\[ \frac{n - 1}{2} = \frac{11 - 1}{2} = 5 \]
For odd n, \(K_n\) does decompose into exactly \((n-1)/2\) edge-disjoint Hamiltonian cycles, so 5 is achieved. Eleven would need 22 neighbours each, and 4 or 7 do not match the bound.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development