Question 1210218: Find the number of ways of filling in the squares of a 3 \times 3 grid so that:
* Each square contains a 0 or a 1.
* The sum of the numbers in each row and each column is at most 1.
An example is shown below.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Solution:
Let the $3 \times 3$ grid be represented by a matrix $A$ where $A_{ij} \in \{0, 1\}$ for $1 \le i, j \le 3$.
The conditions are:
1. Each entry is either 0 or 1.
2. The sum of the numbers in each row is at most 1.
3. The sum of the numbers in each column is at most 1.
Consider the number of 1s in the grid. Let $k$ be the number of 1s.
Since the sum of each row is at most 1, each row can have at most one 1.
Since the sum of each column is at most 1, each column can have at most one 1.
If there are $k$ ones in the grid, these $k$ ones must be placed in $k$ different rows and $k$ different columns.
Case 1: $k = 0$ (No 1s)
If there are no 1s, all entries are 0. This satisfies the conditions. There is $\binom{3}{0} \times \binom{3}{0} = 1$ way to place 0 ones. The grid is $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$. Number of ways = 1.
Case 2: $k = 1$ (One 1)
If there is one 1, it can be placed in any of the $3 \times 3 = 9$ squares.
The row sum and column sum for the row and column containing the 1 will be 1, and for all other rows and columns will be 0. This satisfies the conditions. Number of ways = $\binom{3}{1} \times \binom{3}{1} \times 1! = 3 \times 3 \times 1 = 9$.
Case 3: $k = 2$ (Two 1s)
If there are two 1s, they must be in different rows and different columns.
Choose 2 rows out of 3: $\binom{3}{2} = 3$ ways.
Choose 2 columns out of 3: $\binom{3}{2} = 3$ ways.
Place the two 1s in the chosen $2 \times 2$ subgrid such that each row and each column has one 1. This can be done in $2! = 2$ ways.
Number of ways = $\binom{3}{2} \times \binom{3}{2} \times 2! = 3 \times 3 \times 2 = 18$.
Case 4: $k = 3$ (Three 1s)
If there are three 1s, they must be in different rows and different columns.
Choose 3 rows out of 3: $\binom{3}{3} = 1$ way.
Choose 3 columns out of 3: $\binom{3}{3} = 1$ way.
Place the three 1s in the $3 \times 3$ grid such that each row and each column has one 1. This is equivalent to the number of permutations of 3 elements, which is $3! = 6$ ways.
Number of ways = $\binom{3}{3} \times \binom{3}{3} \times 3! = 1 \times 1 \times 6 = 6$.
Case 5: $k \ge 4$
If there are 4 or more 1s, it is impossible to satisfy the condition that the sum of the numbers in each row and each column is at most 1, because by the Pigeonhole Principle, at least one row or one column would have a sum of at least 2.
The total number of ways is the sum of the number of ways for each case:
Total ways = (ways with 0 ones) + (ways with 1 one) + (ways with 2 ones) + (ways with 3 ones)
Total ways = $1 + 9 + 18 + 6 = 34$.
Final Answer: The final answer is $\boxed{34}$
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