图1显示了一个盘子，图2显示了两叠相同的盘子。较矮的一叠有3个盘子，较高的一叠有7个盘子。较矮一叠的高度为20 cm，较高一叠的高度为44 cm。

1. Identify the difference in height and the difference in the number of plates between the two stacks.

Difference in height $= 44 - 20 = 24$ cm
Difference in number of plates $= 7 - 3 = 4$ plates

2. Calculate the height of one plate gap (the thickness added by each additional plate).

Height of 1 gap $= 24 \div 4 = 6$ cm

3. Calculate the height of a single plate by subtracting the gaps from the total height of a stack.

Height of 3 plates stack $= 20$ cm
This stack consists of 1 base plate and 2 gaps.
Height of 2 gaps $= 2 \times 6 = 12$ cm
Height of 1 plate $= 20 - 12 = 8$ cm

1. Find the height of a single plate and the height added by each subsequent plate in a normal stack.

From the given stacks:
Height of 7 plates = $44$ cm
Height of 3 plates = $20$ cm
Difference in height for 4 plates = $44 - 20 = 24$ cm
Height added by each additional plate in a stack = $24 \div 4 = 6$ cm
Height of the first plate = $20 - 2 \times 6 = 8$ cm

2. Calculate the maximum number of plates that can be packed in Arrangement A.

In Arrangement A, the plates are stacked normally.
Let $n$ be the number of plates.
Total height = $8 + 6(n - 1) = 6n + 2$ cm
We need the height to be at most $1$ m = $100$ cm.
$6n + 2 \le 100$
$6n \le 98$
$n \le 16.33$
So, a maximum of $16$ plates can be packed in Arrangement A.

3. Calculate the maximum number of plates that can be packed in Arrangement B and find the difference.

In Arrangement B, the plates are in an alternating orientation (e.g., top-to-top, base-to-base), meaning they do not overlap and each plate contributes its full height of $8$ cm.
Number of plates in Arrangement B = $100 \div 8 = 12.5$
So, a maximum of $12$ plates can be packed in Arrangement B.
Difference = $16 - 12 = 4$
Arrangement A allows Matthew to pack $4$ more plates.