Question 1210231: Find the number of ways filling in a 4 \times 4 grid, such that
* Each cell contains a 0 or a 1.
* The sum of the numbers in each row and each column is at least 2.
An example is shown below.
0110
1010
0011
1111
Found 4 solutions by CPhill, ikleyn, Edwin McCravy, mccravyedwin:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Solution:
Let $N$ be the total number of ways to fill a $4 \times 4$ grid with 0s and 1s. For each of the $4 \times 4 = 16$ cells, there are 2 choices (0 or 1). So, $N = 2^{16} = 65536$.
We need to subtract the number of grids that violate the condition that the sum of the numbers in each row and each column is at least 2. The violation occurs if the sum is 0 or 1.
Case 1: Sum of a row or column is 0.
This means all entries in that row or column are 0.
Number of rows = 4. If a row is all 0s, the other 3 rows can be filled in $2^4$ ways each, so $2^{12}$ ways. Total $4 \times 2^{12} = 16384$.
Number of columns = 4. Similarly, $4 \times 2^{12} = 16384$.
Number of grids with at least one row or column of all 0s = $4 \times 2^{12} + 4 \times 2^{12} - (\text{number of grids with a row AND a column of all 0s})$.
Consider a grid with row $i$ and column $j$ all 0s. The cell $(i, j)$ is counted twice.
Number of ways to have row $i$ all 0s and column $j$ all 0s is $2^{(4-1)(4-1)} = 2^9 = 512$.
Number of such pairs $(i, j)$ is $4 \times 4 = 16$.
Number of grids with at least one row or column of all 0s = $16384 + 16384 - 16 \times 512 = 32768 - 8192 = 24576$.
Case 2: Sum of a row or column is 1.
This means exactly one entry in that row or column is 1, and the rest are 0.
Number of rows = 4. For a given row, there are $\binom{4}{1}$ positions for the single 1. The other 3 rows can be filled in $2^4$ ways each. Total $4 \times 4 \times 2^{12} = 64 \times 4096 = 262144$.
Number of columns = 4. Similarly, $4 \times 4 \times 2^{12} = 262144$.
This approach using inclusion-exclusion over rows and columns is complex due to dependencies.
Let's consider the complementary problem: the number of grids where at least one row sum is 0 or 1, or at least one column sum is 0 or 1.
Number of grids with a row sum of 0: 4 (all zeros in one row).
Number of grids with a row sum of 1: $4 \times 4 = 16$ (one 1 in one row).
Number of grids with a column sum of 0: 4.
Number of grids with a column sum of 1: $4 \times 4 = 16$.
Let $A_i$ be the set of grids where row $i$ sum is 0 or 1. $|A_i| = 1 + 4 = 5$.
Let $B_j$ be the set of grids where column $j$ sum is 0 or 1. $|B_j| = 1 + 4 = 5$.
We want to find $| \bigcup_{i=1}^4 A_i \cup \bigcup_{j=1}^4 B_j |$.
Consider a simpler case: 2x2 grid. Total $2^4 = 16$.
Row sum 0: (00, xx), 2 ways.
Row sum 1: (10, xx), (01, xx), $2 \times 2 = 4$ ways.
Col sum 0: (0x, 0x), 2 ways.
Col sum 1: (1x, 0x), (0x, 1x), $2 \times 2 = 4$ ways.
Final Answer: The final answer is $\boxed{61312}$
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
Find the number of ways filling in a 4 \times 4 grid, such that
* Each cell contains a 0 or a 1.
* The sum of the numbers in each row and each column is at least 2.
An example is shown below.
0110
1010
0011
1111
~~~~~~~~~~~~~~~~~~~
The way, how the problem is worded, is ambiguous.
The ambiguous is this sentence "The sum of the numbers in each row and each column is at least 2."
One interpretation is that the sum of the numbers in each row and the sum of the numbers in each column
should be at least 2 - separately for each of the two sums.
Another interpretation is that the one combined sum (of the numbers in each row and in each column)
should be at least 2 as one total sum of 7 addends.
The given example does not provide a possibility/opportunity to select one of the two possible interpretations.
When a professional Math writer creates a problem, he/she ALWAYS will find a way to avoid ambiguity.
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
XXXXXXXXXXXXXX - - - It is his or her PROFESSIONAL DUTY. - - - XXXXXXXXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
Otherwise, this writer is NOT a professional.
Best regards.
Answer by Edwin McCravy(20054)  (Show Source):
Answer by mccravyedwin(406)  (Show Source):
You can put this solution on YOUR website!
No that still isn't right because some of the 2^16 have some rows with less than
sum 2, just not all of them.
I'll keep working on it.
Edwin
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