To prove or disprove the given statement:
Given a 2n x n matrix with column sum and row sum equal to 2r and r, where 1 ≤ r < n, we need to show that it is always possible to mark 2n 1's such that exactly one 1 is marked in each row and exactly two 1's are marked in each column.
Proof:
Consider the 2n x n matrix with the given conditions. Let's try to construct the marking as described:
1. Start by marking the 1's in the rows:
- Since each row sum is r, we can mark one 1 in each row to fulfill the condition of exactly one 1 marked in each row.
2. Next, consider the columns:
- Each column sum is 2r, meaning there are two 1's in each column.
- If there are columns with exactly two 1's, we can mark both of them.
- If there are columns with more than two 1's, choose any two of them to mark.
With this approach, we can mark 2n 1's such that exactly one 1 is marked in each row and exactly two 1's are marked in each column.
Counterexample:
If we consider a case where n = 1 and r = 1, the conditions become:
2 x 1 matrix with column sum = 2 and row sum = 1
In this scenario, it is not possible to mark 2 1's such that exactly one 1 is marked in each row and exactly two 1's are marked in each column. This counterexample shows that the statement is not universally true for all values of n and r.
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