The squares of a 3×3 grid are filled with non-negative integers such that the sum of each row and the sum of each column is 7. How many different ways can the squares be filled? The numbers in each grid square does not need to be distinct. Rotations and reflections are distinct arrangements.
To solve this problem, we can use the concept of generating functions. Let's define a polynomial P(x) whose coefficients represent the number of ways to fill the squares. The exponent of each term corresponds to the sum of the numbers in a row or column.
First, let's consider the sum of a row. Since there are 3 squares in a row, the sum will be represented by a polynomial with 3 terms. We can write this polynomial as (1 + x + x^2 + x^3).
Similarly, the sum of a column will also be represented by the same polynomial (1 + x + x^2 + x^3).
Now, we can express the total sum of the grid by multiplying the polynomial for a row with the polynomial for a column. To find the sum of each row and column to be 7, we need the coefficient of x^7 in the product of these two polynomials.
To do this, we can multiply the polynomials using long multiplication or convolution. The resulting polynomial will have 3+3-1 = 5 terms. Writing out the multiplication in full, we get:
P(x) = (1 + x + x^2 + x^3) * (1 + x + x^2 + x^3)
= 1 + 2x + 3x^2 + 4x^3 + 3x^4 + 2x^5 + 1x^6
= 1 + 2x + 3x^2 + 4x^3 + 3x^4 + 2x^5 + x^6 (since x^6 has coefficient 0)
Now, we need to find the coefficient of x^7 in P(x). In this case, it is 0, since there is no term with x^7.
Therefore, there are 0 ways to fill the squares of the 3x3 grid such that the sum of each row and the sum of each column is 7.
Note: If we were asked for the number of ways to fill the grid if the sum of each row and column was 6 instead of 7, we would need to evaluate the coefficient of x^6 in P(x), which is 1.