In how many ways can you break a off a rectangular piece of chocolate from a chocolate bar with m x n squares. [We must respect the structure of the chocolate bar, that is break only along horizontal or vertical lines.
[Math] Number of Ways to Break a Chocolate Bar
combinatorics
Related Solutions
Initially, there is one piece of chocolate. When you have finished, there will be 48. Observe that each break splits a piece into two pieces, and so increases the total number of pieces by 1. To reach 48 pieces, you therefore need 47 breaks, and all methods require this number.
This is a starter providing some ideas which can be used to iteratively determine the number of ways to eat an $(m\times n)$ chocolate bar. We consider an $(m\times n)$ rectangle and start eating from bottom left to top right. The graphic below shows a valid configuration of a $(7\times 4)$ chocolate bar after three bite indicated by $X$.
Valid paths:
We characterize a valid path by an $n$-tupel giving for each $y$, $1\leq y\leq n$ the corresponding $x$-value , $1\leq x\leq m$. The valid path in the graphic is encoded this way as ${(1,2,2,5)}$. We have a total of $\binom{m+n}{n}$ valid paths and consider these paths as building blocks to determine the number of ways to eat the chocolate bar. A valid path is encoded as $(x_1,x_2,\ldots,x_n)$ with $0\leq x_1\leq \cdots \leq x_n\leq m$. The final path is $(m,m,\ldots,m)$.
In order to determine the number of ways to obtain $(1,2,2,5)$ we consider all possible predecessors from which we can get $(1,2,2,5)$ in one step. We add up the number of ways to obtain all predecessors and get so the number of ways for $(1,2,2,5)$. The predecessors of $(1,2,2,5)$ are indicated by the grey shaded regions and are \begin{align*} (\color{blue}{0},2,2,5)\qquad (1,2,2,\color{blue}{2})\\ (1,\color{blue}{1},2,5)\qquad (1,2,2,\color{blue}{3})\\ (1,\color{blue}{1},\color{blue}{1},5)\qquad (1,2,2,\color{blue}{4})\\ \end{align*} The blue marked coordinates are to bite off to come to $(1,2,2,5)$.
Example: $m=n=3$
We determine this way the number $p_{(3,3,3)}$ of possible ways to eat a $(3\times 3)$ chocolate bar which is according to OP's table \begin{align*} \color{blue}{p_{(3,3,3)}=1\,232} \end{align*} We start determining the $\binom{6}{3}=20$ valid paths. These are:
\begin{align*} &(0,0,0)\\ &(0,0,1)\,(0,1,1)\quad\quad\quad\quad\quad\quad\,\,\,\, (1,1,1)\\ &(0,0,2)\,(0,1,2)\,(0,2,2)\qquad\quad\,\,\,(1,1,2)\,(1,2,2)\qquad\quad\,\,\,(2,2,2)\\ &(0,0,3)\,(0,1,3)\,(0,2,3)\,(0,3,3)\,(1,1,3)\,(1,2,3)\,(1,3,3)\,(2,2,3)\,(2,3,3)\,(3,3,3) \end{align*}
We calculate iteratively $p_{(3,3,3)}$ by starting with $p_{(0,0,0)}=1$. We obtain \begin{align*} p_{(0,0,0)}&=1\\ \color{blue}{p_{(0,0,1)}}&=p_{(0,0,0)}\color{blue}{=1}\\ \color{blue}{p_{(0,0,2)}}&=p_{(0,0,1)}+p_{(0,0,0)}=1+1\color{blue}{=2}\\ \color{blue}{p_{(0,0,3)}}&=p_{(0,0,2)}+p_{(0,0,1)}+p_{(0,0,0)}=2+1+1\color{blue}{=4}\\ \\ \color{blue}{p_{(0,1,1)}}&=p_{(0,0,1)}+p_{(0,0,0)}=1+1\color{blue}{=2}\\ p_{(0,1,2)}&=p_{(0,1,1)}+p_{(0,0,1)}+p_{(0,0,0)}=2+1+1=4\\ p_{(0,1,3)}&=p_{(0,1,2)}+p_{(0,1,1)}+p_{(0,0,3)}=4+2+4=10\\ \color{blue}{p_{(0,2,2)}}&=p_{(0,1,2)}+p_{(0,1,1)}+p_{(0,0,2)}\\ &\quad+p_{(0,0,1)}+p_{(0,0,0)}=4+2+2+1+1\color{blue}{=10}\\ p_{(0,2,3)}&=p_{(0,2,2)}+p_{(0,1,3)}+p_{(0,0,3)}=10+10+4=24\\ \color{blue}{p_{(0,3,3)}}&=p_{(0,2,3)}+p_{(0,2,2)}+p_{(0,1,3)}+p_{(0,1,2)}\\ &\quad+p_{(0,1,1)}+p_{(0,0,3)}+p_{(0,0,2)}+p_{(0,0,1)}+p_{(0,0,0)}\\ &=24+10+10+4+2+4+2+1+1\color{blue}{=58}\\ \\ \color{blue}{p_{(1,1,1)}}&=p_{(0,1,1)}+p_{(0,0,1)}+p_{(0,0,0)}=2+1+1\color{blue}{=4}\\ p_{(1,1,2)}&=p_{(1,1,1)}+p_{(0,1,2)}+p_{(0,0,2)}=4+4+2=10\\ p_{(1,2,2)}&=p_{(1,1,2)}+p_{(1,1,1)}+p_{(0,2,2)}=10+4+10=24\\ p_{(1,1,3)}&=p_{(1,1,2)}+p_{(1,1,1)}+p_{(0,1,3)}+p_{(0,0,3)}=10+4+10+4=28\\ p_{(1,2,3)}&=p_{(1,2,2)}+p_{(1,1,3)}+p_{(0,2,3)}=24+28+24=76\\ p_{(1,3,3)}&=p_{(1,2,3)}+p_{(1,2,2)}+p_{(1,1,3)}+p_{(1,1,2)}+p_{(1,1,1)}\\ &=76+24+28+10+4+58=200\\ \\ \color{blue}{p_{(2,2,2)}}&=p_{(1,2,2)}+p_{(1,1,2)}+p_{(0,2,2)}+p_{(0,1,2)}+p_{(0,0,2)}\\ &\quad+p_{(1,1,1)}+p_{(0,1,1)}+p_{(0,0,1)}+p_{(0,0,0)}\\ &=24+10+10+4+2+4+2+1+1\color{blue}{=58}\\ p_{(2,2,3)}&=p_{(2,2,2)}+p_{(1,2,3)}+p_{(1,1,3)}\\ &\quad+p_{(0,2,3)}+p_{(0,1,3)}+p_{(0,0,3)}\\ &=58+76+28+24+10+4=200\\ p_{(2,3,3)}&=p_{(2,2,3)}+p_{(2,2,2)}+p_{(1,3,3)}+p_{(0,3,3)}\\ &=200+58+200+58=516\\ \\ \color{blue}{p_{(3,3,3)}}&=p_{(2,3,3)}+p_{(2,2,3)}+p_{(2,2,2)}+p_{(1,3,3)}+p_{(1,2,3)}\\ &\quad+p_{(1,2,2)}+p_{(1,1,3)}+p_{(1,1,2)}+p_{(1,1,1)}+p_{(0,3,3)}+p_{0,2,3)}\\ &\quad+p_{(0,2,2)}+p_{(0,1,3)}+p_{(0,1,2)}+p_{(0,1,1)}+p_{(0,0,3)}+p_{(0,0,2)}\\ &\quad+p_{(0,0,1)}+p_{(0,0,0)}\\ &=516+200+58+200+76+28+24+10+4+58\\ &\quad+24+10+10+4+2+4+2+1+1\\ &\,\,\color{blue}{=1\,232} \end{align*} and we obtain $p_{(3,3,3)}=1\,232$ according to OP's table. Entries with a rectangular shape are marked in blue. They are also given in OP's list.
Best Answer
HINT: An $m\times n$ bar has $m+1$ division lines in one direction, say horizontally, and $n+1$ in the other. To determine a rectangle, pick two horizontal and two vertical division lines. In how many ways can you do this?