Problem on Permutation | SMO, 2011 | Problem No. 24

If you really get stuck in this problem here is the first hint to do that:

At 1st let’s consider the sub grids of \( 2 \times 2\) filled with 1-4 ( 1, 2 , 3 ,4)

If a,b,c,d are all distinct , and there are no other numbers to place in x . If {a,b} = {c,d} then again a’,b’,c,d are all distinct , and no other number can be possible for x’.

We need to understand that the choices we have ,

{a,a’} = {1,2} , {b,b’} = {3,4}, {c,c’} = {2,4} and {d,d’} = {1,3}

Among these choices \( 2^4 = 16 \) choices 4 of them are impossible – {a,b} = {c,d} = {1,4} or {2,3} and

{a,b} = {1,4} and {c,d} = {2,3} and {a,b} = {2,3} and {c,d} = {1,4}

Try rest….

Now for each remaining case a’,b’,c’ and d’ are uniquely determined so

{x} = {1,2,3,4} – {a,b} \(\cup\) {c,d}

{y} = {1,2,3,4} – {a,b} \(\cup\) {c’,d’}

{x’} = {1,2,3,4} – {a’,b’} \(\cup\) {c,d}

{y’} = {1,2,3,4} – {a’,b’} \(\cup\) {c’,d’}

In final hint :

There are 4! = 24 permutation in the left top grid we can find. So total 12 * 24 = 288 possible 4\(\times\) 4 Sudoku grids can be found.