Consider the following 4 × 4 matrix,
7 18 11 9 13 24 17 15 3 14 7 5 10 21 14 12
and the following procedure applied to the matrix:
Step 1. Step 2.
Step 3. Step 4.
Pick any number in the matrix.
Pick any other number in the marix that is not in the row or column the first number is in.
Pick any other number in the matrix that is not in a row or column that the first or second number is in.
Pick the number in the matrix that is not in a row or column that the first, second, or third number. is in.
It turns out that the four numbers selected in the process, no matter what choices were made within the rules, will add up to 50. Your task, should you choose to accept it, is to answer the following question:
1. Why does the given procedure, when applied to the given matrix, always select four number that add up to 50? [10]

The given procedure involves picking four numbers from a \(4 \times 4\) matrix, following specific rules:

1. Pick any number from the matrix.
2. Pick another number that is **not** in the same row or column as the first number.
3. Pick another number that is **not** in the same row or column as the first or second number.
4. Pick the final number that is **not** in the same row or column as the first, second, or third numbers.

We are tasked with explaining why the sum of these four selected numbers is always 50.

### Matrix Setup
The \(4 \times 4\) matrix is as follows:

\[
\begin{bmatrix}
7  & 18 & 11 & 9  \\
13 & 24 & 17 & 15 \\
3  & 14 & 7  & 5  \\
10 & 21 & 14 & 12
\end{bmatrix}
\]

### Analysis of the Procedure

When following the procedure, you are essentially choosing exactly one number from each **row** and exactly one number from each **column**. The reason is that each new number selected must be in a different row and column from all previously selected numbers.

This process effectively forms a **Latin square** pattern on the matrix, where each number chosen comes from a different row and a different column.

### Key Observation

Since you are always choosing one number from each row and column, the four selected numbers will correspond to one permutation of a particular set of entries in the matrix. 

For example, let’s look at the sum of all elements in **any fixed row and column position** across the entire matrix:

\[
\begin{aligned}
(7, 24, 7, 12) \quad &\text{(choosing from top-left to bottom-right diagonal)} \\
(18, 17, 5, 10) \quad &\text{(other choices based on row/column exclusion)}
\end{aligned}
\]

Summ !

Consider the following 4 × 4 matrix: 7 18 11 9 13 24 17 15 3 14 7 5 10 21 14 12, and the following procedure applied to the matrix: Step 1. Pick any number in the matrix. Step 2. Pick any other number in the matrix that is not in the row or column the first number is in. Step 3. Pick any other number in the matrix that is not in a row or column that the first or second number is in. Step 4. Pick the number in the matrix that is not in a row or column that the first, second, or third number is in. It turns out that the four numbers selected in the process, no matter what choices were made within the rules, will add up to 50. Why does the given procedure, when applied to the given matrix, always select four numbers that add up to 50?

Explore why the sum of four selected numbers from a 4x4 matrix always adds up to 50, based on a step-by-step selection procedure. This problem introduces concepts of matrix theory and combinatorics, along with the properties of Latin squares. Suitable for students in Grades 10-12.

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