In how many ways can the following words be read from the diagrams: TISZA (1st diagram), CSERE, BERE, CSEREBERE (2nd diagram), if

a) it is only allowed to step from any letter to an adjacent letter;

b) any letter can only be used once in a single reading?

| $T$ | $I$ | $S$ | $Z$ | $A$ |
| :---: | :---: | :---: | :---: | :---: |
| $I$ | $S$ | $Z$ | $A$ |  |
| $S$ | $Z$ | $A$ |  |  |
| $Z$ | $A$ |  |  |  |
| $A$ |  |  |  |  |

| $C$ | $S$ | $E$ | $R$ | $E$ | $B$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $S$ | $E$ | $R$ | $E$ | $B$ | $E$ |
| $E$ | $R$ | $E$ | $B$ | $E$ | $R$ |
| $R$ | $E$ | $B$ | $E$ | $R$ | $E$ |
| $E$ | $B$ | $E$ | $R$ | $E$ | $S$ |
| $B$ | $E$ | $R$ | $E$ | $S$ | $C$ |

1st diagram
2nd diagram
After reaching any of the letters $T, I, S, Z$, the next letter of the word can be chosen from 2 neighbors, meaning the reading can branch out in two ways at each letter. The letter $T$ appears once, so the word $T I S Z A$ can be read in $1 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 16$ ways.

To provide a solution for further questions, we also give a method that can be used for reading $TISZA$ in structures of different shapes. Write the number of ways to reach each letter of part a) of the diagram in place of the corresponding letter in part b), starting from $T$. For example, there is 1 way to reach $T$, 1 way to reach each $I$, 1 way to reach each outer $S$, 2 ways to reach the middle $S$ (via the two $I$s), and so on. Similarly, there are $1+2$ ways to reach each inner $Z$. The desired number is given by the sum of the numbers written at the $A$s.

![](https://cdn.mathpix.com/cropped/2024_05_02_411e4e93bb3a15aacfc1g-1.jpg?height=268&width=1410&top_left_y=511&top_left_x=346)

Part c) of the diagram also answers in this way, but based on the somewhat peculiar observation: the number of ways to read $T I S Z A$ is the same as the number of ways to read $A Z S I T$ in reverse. The numbers in corresponding positions in parts b) and c) of the diagram - as shown in part d) - also give the number of readings that use each letter; for example, the $Z$ in the second column has $3 \cdot 2$ paths: 3 types of $T I S Z$ and 2 types of $A Z$ arrive at it, and each type of $T I S Z$ can be connected with each type of $Z A$. (The sum of the numbers in each diagonal row of the same type of letter gives the total number of readings.)

In reading the words $CSERE$ and $BERE$, a turn is possible at the $R$ in $CSEREBERE$, and at the $B$ in $BERE$; that is, after moving down and to the right, the last $E$ or the $B E R E$ segment can also be taken by moving up or to the left. The $CSEREBERE$ diagram is mapped onto itself by reflection in any of its diagonals, so it is enough to count the readings that start from the upper $C$ and use one of the $R$s in the 1st and 2nd columns; the desired number is clearly four times this. One and three paths lead to the mentioned $R$s, and the number of adjacent $E$s is 3 and 4, respectively, and we can move to all but one of them, i.e., the one we arrived from; thus, the incoming paths can be completed in 2 and 3 ways, respectively. The number of readings - including the symmetric cases: $4(1 \cdot 2 + 3 \cdot 3) = 44$ (with the idea of part d)).

The $BERE$ case differs from this only in that 4 paths lead to each $R$ - part c) - so the number of readings is $4(4 \cdot 2 + 4 \cdot 3) = 80$.
![](https://cdn.mathpix.com/cropped/2024_05_02_411e4e93bb3a15aacfc1g-1.jpg?height=298&width=938&top_left_y=1504&top_left_x=539)

In the word $CSEREBERE$, a turn is only possible at the second $R$ and at the $B$. The total number of readings without turning and with turning only at the second $R$ is, following the pattern of $CSERE$, $4(21 \cdot 2 + 35 \cdot 3) = 588$. - To count the total number of readings with a turn at $B$, the different possibilities must be examined in more detail. It is enough to consider the $B$s in the 2nd and 3rd columns, as a turn is not possible at the corner $B$s. Of the 5 paths from the upper $C$ to the second $B$, 2 do not have a turning $BERE$ ending, and the other 3 have 1 and 2 endings, respectively, totaling 4. - To the third $B$, 4 paths arrive from the left and 6 from above. The paths arriving with a single direction change provide the most possibilities for the turning $BERE$ reading, 4 and 3, respectively. It is easy to count that 4, 3, 2, and 2 endings correspond to the paths arriving from the left, and 3, 2, 1, 2, 1, and 0 to those arriving from above, totaling 20, so the total number of $CSEREBERE$ readings with a turn at $B$ is $4(0 + 4 + 20) = 96$, and there are only 684 reading possibilities.