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In the 3rd section titled : 'Counting Solutions' for the webpage here, there is calculation as shown here that has exercise based on filling the $9$ sub-cubes of Sudoku named as $B_1, B_2, B_3$ in the first row, & so on as shown in diagram below. $$ \left[ \begin{array}{c|c|c} B_1&B_2&B_3\\ -&-&-\\ B_4&B_5&B_6\\ -&-&-\\ B_7&B_8&B_9\\ \end{array} \right] $$

Exercise:- List all the possible ways of filling in the first rows in B2 and B3, up to reordering of the digits in each block. $$ \left[ \begin{array}{ccc|ccc|ccc} 1&2&3&*&*&*&*&*&*\\ 4&5&6&*&*&*&*&*&*\\ 7&8&9&*&*&*&*&*&*\\ -&-&-&-&-&-&-&-&-\\ *&*&*&*&*&*&*&*&*\\ *&*&*&*&*&*&*&*&*\\ *&*&*&*&*&*&*&*&*\\ -&-&-&-&-&-&-&-&-\\ *&*&*&*&*&*&*&*&*\\ *&*&*&*&*&*&*&*&*\\ *&*&*&*&*&*&*&*&*\\ \end{array} \right] $$

It states as hint: There are ten ways of doing this so that swapping B2 and B3 in these ten ways give you ten more ways, for a total of twenty.

I am unable to get figure of $10$ ways & request help.
Let, $B_2$ be the first to be filled. So, selects $3$ elements out of $6$ left. Say, the first row of $B_2$ has selected $3$ elements: $4,5,6$. This leads to $^6P_3(=120)$ ways for the selected $3$ elements of the first row of $B_2 $.
These $3$ elements then need be reordered to get $^3P_3(=3!=6)$ further ways.
Applying the product rule, get $720$ ways.
For the first row of $B_2$ get further $3!$ arrangements. The multiplication leads to $720*3= 2160$ ways.

Even if I ignore the ways of selecting $3$ elements from $6$, still get $3!*3!= 36$ ways.

An unrelated but quite imp. question on shidoku, on mse here. It helps to find impossible cases for a $4\times 4$ block.

jiten
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1 Answers1

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It says "up to reordering of the digits in each block".

So, for example, we consider $$1\ 2\ 3\mid 4\ 5\ 6\mid 7\ 8\ 9$$ is the same as $$1\ 2\ 3\mid 5\ 6\ 4\mid 7\ 8\ 9$$

So, the total number is given by $\binom{6}{3}=20$.

mathlove
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  • Thanks a lot. I am unable to get an approach to directly get $10$ for the sub-cube $B_2$ or $B_3$. The only way to get is to divide $\binom{6}{3}=20$ by $2$. – jiten Oct 02 '18 at 06:08
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    @jiten: I'm not very sure what the hint is saying, but one way is as follows : we consider first the number of the possible ways such that $4$ is in $B_2$. The number is $\binom{5}{2}=10$. So, the total number is $10\times 2$. – mathlove Oct 02 '18 at 06:18
  • So, choosing (combinations) of $3$ elements from $6$ is the same as taking one element as fixed for $B_2$ & then getting the no. of choices (of filling $2$ elements from $5$ available); & multiplying this by $2$ (for which ever cube gets filled first) for getting the total number of choices available. – jiten Oct 02 '18 at 06:27
  • I hope that the page#$1$ at : http://www.afjarvis.staff.shef.ac.uk/sudoku/sudoku.pdf shows the set of $10$ values for each $B_2, B_3$; & their swap leading to a total of $20$ choices. – jiten Oct 03 '18 at 07:05