Answer:
(B) 6
Explanation:
You want the smallest number of cells in a 5×5 matrix that can be colored so that there are no 4 consecutive uncolored cells in any row or column.
Solution
The minimal coloring can be done any of 18 ways. Here is one of them:
![\left[\begin{array}{ccccc}\cdot&\square&\cdot&\cdot&\cdot\\\square&\cdot&\cdot&\cdot&\square\\\cdot&\cdot&\square&\cdot&\cdot\\\cdot&\cdot&\cdot&\square&\cdot\\\cdot&\square&\cdot&\cdot&\cdot\end{array}\right]](https://img.qammunity.org/2024/formulas/mathematics/college/c9r073hmjkjvsnojw01uzvokaak62vm581.png)
Where each square represents a colored cell.
At least one cell on each edge must be colored (4). If those are arranged so that cells at each end of a row or column are colored, then that leaves two rows and two columns with no coloring. Two (2) additional cells are required to break up those into sections of 3 or less. That means a total of 6 cells must be colored to achieve the desired condition.