A person divides a square to 100 small squares with equal sizes. Each small square is randomly selected and given a number from 1-100, such that each square is 'unique'.
Then the person counts the sum of the numbers at each row, column, and each diagonal. In total, 22 numbers (22 sum results).
What is the probability such that the average of those 22 numbers is the smallest possible average..?
Attempt:
From my understanding, the diagonals are the two diagonals, and they certainly do not overlap each other at the center. Thisis because the size is 100 (even number).
The placing of numbers on the rows and columns doesnt affect the total values. Any permutation will have a total $5050$.
So the average will be affected only by the permutations in the two diagonals. The smallest average is when the numbers $1-20$ are in the diagonals. Hence the number of outcomes (the permutation) is $$ (20 \times 19 \times ..... \times 2 \times 1) = 20! $$
And it should be multiplied by $$ 80! $$ Which is the number of placing for the non-diagonals.
While the total possible outcome should be
$$ 100! $$
So the probability should be :
$$ \frac{80! 20!}{100!} $$
Is this accurate? Are there better methods? Thanks.