To solve the problem of finding the number of valid ways to fill a 2x3 grid with 0s and 1s such that each row has exactly two 1s and every column has at least one 1, follow these steps:

Step 1: Total possible rows

Each row must have exactly two 1s. For a 3-column row, the number of ways to choose 2 positions for 1s is $\binom{3}{2}=3$. Let these row types be:

• A: [1,1,0] (columns 1&2 are 1)
• B: [1,0,1] (columns 1&3 are 1)
• C: [0,1,1] (columns 2&3 are 1)

Step 2: Total pairs of rows

Since there are 2 rows, total possible pairs are $3 \times 3 = 9$ (each row has 3 choices).

Step 3: Identify invalid pairs

A pair is invalid if any column has zero 1s:

• AA: column 3 is zero → invalid
• BB: column 2 is zero → invalid
• CC: column 1 is zero → invalid

Total invalid pairs: 3.

Step 4: Valid pairs

Valid pairs are those where all columns have at least one 1. These are the cross pairs of different row types:
(A,B), (A,C), (B,A), (B,C), (C,A), (C,B) → 6 valid pairs.

Answer: $\boxed{6}$

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