Considering Sudoku 3

Probability Level 5

Using the general principles and goals of a Sudoku puzzle there are many thousands of possible arrangements of digits that would satisfy the conditions for a Sudoku solution.

In order to maintain the possibility of a proper Sudoku solution still available in the grid. Considering only the digits in one column and one row. How many possible different arrangements of digits could be displayed if one full column as well as one full row are filled with digits? i.e. how many ways to replace the “?”s in the grid shown?

The answer is 7838208000.

2 solutions

Jaleb Jay
Jan 12, 2016

First, notice that the top left 3x3 box has the greatest restriction of numbers of 5, with none repeating. With this you need to choose 5 of the 9 numbers in any orientation or $(9)_5$ . Next, with the left column already using 3 values, the remaining 6 can be organized in any order or 6!. Similarly, the same can be said about the top row.

Multiplying these values together we get: $(9)_56!6! = \frac{9!6!6!}{(9-5)!} = 7838208000$

Darryl Dennis
Jan 11, 2016

The duplication of any digit in a row, column or block is not allowed in a valid Sudoku solution. To fill any cell with a digit that will maintain a valid solution all of the digits that are currently in the same row, same column, or same block must be excluded from the selection of digits available. Since all possible duplication of a digit in the row, column and blocks have been removed from the eligible possibilities a valid solution is certain with all permutations of the remaining (eligible) digits.

In the first column there are 9 possible digits in total and 9 locations where they can by placed. 9! = 362,880 possible different arrangements in column A

Considering the top row, there are three digits already in place in the left block which leaves 9-3 =6 remaining for the first cell in the top row then five for the last cell in the top row for that block.

5 X 6 =30 possible different arrangements in the left block

Considering the other cell in the top row. there are now 3 cells filled in the top row leaving 6! ways to arrange the remaining 6 digits into the remaining six top row cells.

6!=720 possible different arrangements in the left 6 cells of the top row

total = 9! X 6! X 30 = 7,838,208,000

You haven't proven that each of those possible arrangements leads to a valid Sudoku. (That is, some of the arrangements might actually not give a solvable Sudoku, and hence shouldn't be counted.)

Ivan Koswara - 5 years, 5 months ago

I would like to see some arrangement that cannot lead to a possible Sudoku solution, if you could show one. or at least explain. Each cell can only contain digits that are reaming in a valid solution how is it possible to not be valid?

Darryl Dennis - 5 years, 5 months ago

Well, that's the point; I myself am pretty sure all of them will lead to a solution, but I haven't proved it, and neither has your solution, so your solution is at best incomplete.

Ivan Koswara - 5 years, 5 months ago

@Ivan Koswara – I will explain a more complete solution later I have to go now. basically the digits that are matching a digits in the same row, column, or block are excluded from the selection for each sell.

Darryl Dennis - 5 years, 5 months ago

@Darryl Dennis – I'm trying to think of the best way to prove this as you only need 3 cases to examine.

Jaleb Jay - 5 years, 5 months ago

@Jaleb Jay – I would expect a typical advanced/expert level puzzle might also be enough of an example if you search around.

Jaleb Jay - 5 years, 5 months ago

Define a string of 9 numbers by the first row that is filled in. Fill in each row after by writing the string starting at the number in the first column of that row. When you run out of numbers in the string, go to the first one again. This method will successfully finish any valid combination described in the solution, thus they are all valid.

Austin Antonacci - 5 years, 4 months ago

Considering Sudoku 3

The answer is 7838208000.

2 solutions

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