Key Concepts and Formulas

• Permutation Matrix: A square matrix where each row and each column contains exactly one '1' and the rest are '0's.
• Permutations: The number of ways to arrange n distinct objects in n distinct positions is n! (n factorial), where $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Fundamental Principle of Counting (Multiplication Principle): If there are m ways to do one thing and n ways to do another, then there are $m \times n$ ways to do both.

Step-by-Step Solution

Step 1: Placing the '1' in the First Row

• We need to place a single '1' in the first row of the $5 \times 5$ matrix.
• Since there are no constraints yet from previous rows or columns, we can place the '1' in any of the 5 columns.
• Reasoning: The sum of the elements in the first row must be 1, and the sum of elements in each column must also be 1. Thus, we have 5 choices for the column in which to place the '1'.
• Number of choices for the first row: 5.

Step 2: Placing the '1' in the Second Row

• We now need to place a single '1' in the second row.
• However, we cannot place the '1' in the same column as the '1' in the first row, because that column would then have a sum of 2, violating the column sum constraint.
• Reasoning: To satisfy the condition that the sum of elements in each column is 1, we must choose a different column than the one used in the first row.
• Number of available columns: $5 - 1 = 4$.
• Number of choices for the second row: 4.

Step 3: Placing the '1' in the Third Row

• We need to place a single '1' in the third row.
• We cannot place the '1' in either of the columns that contain the '1's from the first and second rows.
• Reasoning: To satisfy the column sum constraint, the '1' in the third row must be in a column different from those already used.
• Number of available columns: $5 - 2 = 3$.
• Number of choices for the third row: 3.

Step 4: Placing the '1' in the Fourth Row

• We need to place a single '1' in the fourth row.
• We cannot place the '1' in any of the three columns that already have a '1' in them from the first three rows.
• Reasoning: The column sum constraint requires us to choose from the remaining available columns.
• Number of available columns: $5 - 3 = 2$.
• Number of choices for the fourth row: 2.

Step 5: Placing the '1' in the Fifth Row

• We need to place a single '1' in the fifth row.
• There are four columns that already have a '1' in them, leaving only one remaining column.
• Reasoning: To fulfill the column sum constraint, the '1' must be placed in the last remaining column.
• Number of available columns: $5 - 4 = 1$.
• Number of choices for the fifth row: 1.

Step 6: Calculating the Total Number of Matrices

• By the Fundamental Principle of Counting, the total number of such matrices is the product of the number of choices for each row.
• Total number of matrices $= 5 \times 4 \times 3 \times 2 \times 1 = 5!$.
• $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Common Mistakes & Tips

• Confusing with Combinations: The order of column selection matters, so it's a permutation, not a combination. Selecting column 1 then column 2 is different from selecting column 2 then column 1.
• Ignoring Column Constraint: It's crucial to remember that the column sums must also be 1. This limits the choices for subsequent rows.
• Recognize Permutation Matrices: Look for patterns of "exactly one element per row/column" to quickly identify permutation matrix problems, where the answer for an $n \times n$ matrix will be $n!$.

Summary

The problem asks for the number of $5 \times 5$ matrices with entries from {0, 1} such that each row and each column sums to 1. These matrices are permutation matrices. By systematically placing a single '1' in each row, while respecting the column sum constraint, we found that the number of choices for each row decreases. This leads to the expression $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$. Therefore, there are 120 such matrices.

Final Answer

The final answer is \boxed{120}, which corresponds to option (D).