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.
Here's a lower bound solution. You may want to investigate other possibilities.
The number 7 can be partitioned into 3 ascending non-negative integers in 8 ways, namely:
007#
016
025
034
115#
124
133#
233#
There are 3 permutations of partitions (indicated #) with non-distinct digits and 6 permutations of those with distinct digits for a total of
4*3+4*6=36 permutations of non-negative integers whose sum is 7.
Each of these permutations can make a 3x3 grid as follows by putting a permutation as the first line, then rotate the digits to the left:
124
241
412
We can also make another grid by rotating them to the right, as follows:
124
412
241
This doubles the number of grids for a total of
36*2=72 grids.
what about this case, i didn't enumerated it
1 2 4
0 5 2
6 0 1
Hint: you need these:
(1) a,b,c,d are non-negative integers
(2) 7−a−b≥0⇔a+b≤7
(3) 7−c−d≥0⇔c+d≤7
(4) 7−a−c≥0⇔a+c≤7
(5) 7−b−d≥0⇔b+d≤7
(6) a+b+c+d−7≥0⇔a+b+c+d≥7
Try to count how many satisfies these conditions (:
To find the number of different ways the squares can be filled, we can break down the problem into smaller parts.
First, let's look at the constraints given:
1. The sum of each row is 7.
2. The sum of each column is 7.
Let's start with the first constraint.
Step 1: We need to find the number of ways to divide 7 non-negative integers into three groups, which represent the sum of each row of the grid.
To do this, we can use a method known as "Stars and Bars" or "Balls and Urns." This method involves representing the non-negative integers as "stars" and representing the dividing lines between the groups as "bars."
For example, if we have 7 non-negative integers and 2 bars, one possible arrangement could be:
***|**|*
This arrangement represents the first row having 3 numbers, the second row having 2 numbers, and the third row having 1 number.
Step 2: Finding the number of possible arrangements.
To find the number of possible arrangements, we need to find the number of ways to place the bars among the stars. In this case, we have 7 stars and 2 bars.
The formula to find the number of ways to place the bars is given by:
(Number of stars + Number of bars)! / (Number of stars! * Number of bars!)
In this case, it would be:
(7 + 2)! / (7! * 2!) = 9! / (7! * 2!)
Simplifying, we get:
(9 * 8 * 7!) / (7! * 2) = 9 * 4 = 36
Therefore, there are 36 possible arrangements for the sum of each row in the grid.
Step 3: Considering the second constraint.
Now, let's consider the second constraint, which states that the sum of each column is 7.
Since the grid is symmetric, if we know the sum of the first column, we automatically know the sum of the third column. So, we only need to consider the sum of the first column.
The problem now becomes finding the number of ways to divide 7 non-negative integers into three groups, which represent the sum of each column.
Similar to Step 1, we can use the "Stars and Bars" method to represent the non-negative integers as "stars" and the dividing lines between the groups as "bars."
Step 4: Finding the number of possible arrangements.
Using the same formula as Step 2, we can calculate the number of possible arrangements for the sum of each column in the grid.
Again, we have 7 stars and 2 bars, so the calculation would be the same:
(7 + 2)! / (7! * 2!) = 9! / (7! * 2!) = 9 * 4 = 36
So, there are also 36 possible arrangements for the sum of each column in the grid.
Step 5: Combining the arrangements.
Since each row and each column can be filled independently, we can combine the arrangements for the rows and the columns to find the total number of ways the squares can be filled.
The total number of ways is given by multiplying the number of row arrangements by the number of column arrangements:
36 * 36 = 1296
Therefore, there are 1296 different ways the squares can be filled in the 3x3 grid, satisfying the given conditions.