7. In how many ways can we write the word TUESDAY from adjacent fields in the table? One of the ways is shaded in the picture.

|  |  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |
|  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |
| $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{T}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |
| $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{T}$ | $\mathrm{U}$ | $\mathrm{T}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ |
| $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{T}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |
|  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |
|  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |
|  |  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |  |

## Ministry of Science and Education of the Republic of Croatia  Agency for Education and Upbringing  Croatian Mathematical Society

## SCHOOL/CITY COMPETITION IN MATHEMATICS  February 15, 2022  7th grade - elementary school

In addition to the final result, the procedure is also graded. To earn full points, it is necessary to find all solutions and determine that there are no others, write down the procedure, and justify your conclusions. Problems 1. - 5. are worth six points each, and 6. and 7. are worth ten points each.

## Problems worth 6 points:
First solution.

|  |  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
|  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |
|  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |
| $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{T}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |
| $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{T}$ | $\mathrm{U}$ | $\mathrm{T}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ |
| $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{T}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |
|  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{O}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |
|  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{R}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |
|  |  |  | $\mathrm{K}$ | $\mathrm{A}$ | $\mathrm{K}$ |  |  |  |

Let's start from the letter U and move to the letter T. Count how many ways we can get from the letter T to each highlighted letter $\mathrm{K}$.

1 POINT

We can reach the letter $\mathrm{K}$ in the same column as the letter $\mathrm{T}$ in 1 way.

1 POINT

From the letter $\mathrm{T}$ to the next letter $\mathrm{K}$, we can go in 4 ways.

From the letter $\mathrm{T}$ to the next letter $\mathrm{K}$, we can go in 6 ways.

2 POINTS

From the letter $\mathrm{T}$ to the letter $\mathrm{K}$ in the last column, we can go in 4 ways.

1 POINT

Due to the symmetry of the table, from the same letter $\mathrm{T}$ to the letter $\mathrm{K}$ in the lower part of the table, we can also go in $1+4+6+4=15$ ways.

We can conclude that from the letter T to the letter $\mathrm{K}$, we can go in a total of $15+15=30$ ways.

1 POINT

Since we have 4 letters T, there are a total of $4 \cdot 30=120$ ways to spell the word UTORAK.

1 POINT

TOTAL 10 POINTS