Problem (Combinatorics):
An ARMLbar is a $7 \times 7$ grid of unit squares with the center unit square removed. A portion of an ARMLbar is a square section of the bar, cut along the gridlines of the original bar. Compute the number of different ways there are to cut a single portion from an ARMLbar.

Solution:
Note that any portion of side length $m \geq 4$ will overlap the center square, so consider only portions of side length 3 or less. If there were no hole in the candy bar, the number of portions could be counted by conditioning on the possible location of the upper-left corner of the portion. If the portion is of size $1 \times 1$, then the corner can occupy any of the $7^{2}$ squares of the bar. If the portion is of size $2 \times 2$, then the corner can occupy any of the top 6 rows and any of the left 6 columns, for $6^{2}=36$ possible $2 \times 2$ portions. In general, the upper-left corner of an $m \times m$ portion can occupy any of the top $8-m$ rows and any of the left $8-m$ columns. So the total number of portions from an intact bar would be $7^{2}+6^{2}+5^{2}$. Now when $m \leq 3$, the number of $m \times m$ portions that include the missing square is simply $m^{2}$, because the missing square could be any square of the portion. So the net number of portions is

$$
\begin{aligned}
7^{2}+6^{2}+5^{2}-3^{2}-2^{2}-1^{2} & =(49+36+25)-(9+4+1) \\
& =110-14 \\
& =\mathbf{9 6}
\end{aligned}
$$

First ignore the missing square. As in the previous solution, the number of $m \times m$ portions that can fit in the bar is $(8-m)^{2}$. So the total number of portions of all sizes is simply

$$
7^{2}+6^{2}+\cdots+1^{2}=\frac{7(7+1)(2 \cdot 7+1)}{6}=140
$$

To exclude portions that overlap the missing center square, it is useful to consider the location of the missing square within the portion. If an $m \times m$ portion includes the missing center



square, and $m \leq 4$, then the missing square could be any one of the $m^{2}$ squares in the portion. If $m=5$, then the missing square cannot be in the leftmost or rightmost columns of the portion, because then the entire bar would have to extend at least four squares past the hole, and it only extends three. By similar logic, the square cannot be in the top or bottom rows of the portion. So for $m=5$, there are $3 \cdot 3=9$ possible positions. For $m=6$, the two left and two right columns are excluded, as are the two top and the two bottom rows, for $2 \cdot 2=4$ possible positions for the portion. And in a $7 \times 7$ square, the only possible location of the hole is in the center. So the total number of portions overlapping the missing square is

$$
1^{2}+2^{2}+3^{2}+4^{2}+3^{2}+2^{2}+1^{2}=44 .
$$

The difference is thus $140-44=\mathbf{9 6}$

Final answer: 96