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The figure shows a grid formed by a collection of unit squares. The unshaded unit square in the grid represents a hole.

What is the maximum number of squares without a “hole in the interior” that can be formed within the $4 \times 4$ grid using the unit squares as building blocks?

- $15$
- $20$
- $21$
- $26$

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Given that,

If there is no hole in the $4 \times 4$ grid, then the number of square $ = 1^{2} + 2^{2} + 3^{2} + 4^{2} = 1+4+9+16 = 30$

Now, we can calculate, because of the hole, how many squares are not formed.

- $1 \times 1 = 1$ square
- $2 \times 2 = 4$ squares
- $3 \times 3 = 4$ squares
- $4 \times 4 = 1$ square

$\therefore$ The maximum number of squares without a “hole in the interior” that can be formed within the $4 \times 4$ grid using the unit squares as building blocks $ = 30 – 10 = 20$ squares.

Correct Answer $:\text{B}$

$\textbf{PS:}$ The total number of squares in a $ { n\times n \;\text{grid} = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} =\frac{n(n+1)(2n+1)}{6}}.$