Baltic Way 1995 · Problem 11
Combinatorics
In how many ways can the set of integers be partitioned into three nonempty sets so that none of these sets contains two consecutive integers?
When you’re ready
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Review
Topics
Colorings and configurations · Pigeonhole and extremal arguments · Induction and recursion
Solutions
Solution
Solution:
We construct the three subsets by adding the numbers successively, and disregard at first the condition that the sets must be non-empty. The numbers and must belong to two different subsets, say and . We then have two choices for each of the numbers , and different choices lead to different partitions. Hence there are such partitions, one of which has an empty part. The number of partitions satisfying the requirements of the problem is therefore .
Contest context
Results from Baltic Way 1995
9 teams
- Mean score
- 3.9 / 5
- Scores of 4 or 5
- 6 / 9
- Estonia
- 0 / 5
Score distribution
All team scores
| Team | Score |
|---|---|
| Poland | 4 / 5 |
| Latvia | 5 / 5 |
| Sweden | 5 / 5 |
| Lithuania | 3 / 5 |
| Denmark | 5 / 5 |
| Finland | 5 / 5 |
| St. Petersburg | 3 / 5 |
| Estonia | 0 / 5 |
| Iceland | 5 / 5 |